Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Henry Wilson DSc on 8 Mar 2010 15:33 On Mon, 08 Mar 2010 21:07:18 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >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? >============================================================ Your question is nonsense. That which you are refering to is NOT a 'WAVELENGTH'. ....."Never walk down the aisle of an airborne plane"____PA Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 8 Mar 2010 16:04 On 08.03.2010 21:33, Henry Wilson DSc wrote: > On Mon, 08 Mar 2010 21:07:18 +0100, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: >> [snip what's repeated below] >> >> QUESTION #2: >> ============================================================= >> Now the question is: >> How many wavelengths are there in the wave when t>= T? >> Is it 2 or is it 3? >> ============================================================ > > Your question is nonsense. > > That which you are refering to is NOT a 'WAVELENGTH'. Could you be a bit more specific, please? If anything in the following is NOT correct according to the emission theory, please state exactly what is wrong. That is, answer question #1. 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: Henry Wilson DSc on 8 Mar 2010 16:15 On Mon, 08 Mar 2010 22:04:08 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 08.03.2010 21:33, Henry Wilson DSc wrote: >> On Mon, 08 Mar 2010 21:07:18 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >>> [snip what's repeated below] >>> >>> QUESTION #2: >>> ============================================================= >>> Now the question is: >>> How many wavelengths are there in the wave when t>= T? >>> Is it 2 or is it 3? >>> ============================================================ >> >> Your question is nonsense. >> >> That which you are refering to is NOT a 'WAVELENGTH'. > >Could you be a bit more specific, please? > >If anything in the following is NOT correct according to >the emission theory, please state exactly what is wrong. >That is, answer question #1. > >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? >============================================================ There are two 'wavelengths' in the source frame no matter how many differently moving observers look at it. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 9 Mar 2010 15:35 On 08.03.2010 22:15, Henry Wilson DSc wrote: > On Mon, 08 Mar 2010 22:04:08 +0100, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: >> If anything in the following is NOT correct according to >> the emission theory, please state exactly what is wrong. >> That is, answer question #1. >> >> 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. OK. Since you haven't said otherwise, I will assume that you agree that the following is a correct description 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. >> >> (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? >> ============================================================ > > There are two 'wavelengths' in the source frame no matter how many differently > moving observers look at it. Thank you for a clear answer. So we agree: The number of wavelengths in the wave is always two when t > T. The path length of the wave front measured in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. Now we repeat the scenario with the source at the right end of the rod, so the wave going in the other direction. | | |-----------* 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 left end of the rod at x = X : |** ** | *--*--*--*--* phi = 4 pi | ** **| |-----|-----|-----------------------> x 0 X L 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 X L 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/2, so the path length is L/2 or one wavelength. ============================================================= Now the question is: How many wavelengths are there in the wave when t >= T? Is it 2 or is it 1? ============================================================ The answer is obvious. When t > T, there will always be 2 wavelengths in the wave. Don't you agree? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 9 Mar 2010 16:53
On Tue, 09 Mar 2010 21:35:21 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 08.03.2010 22:15, Henry Wilson DSc wrote: >> On Mon, 08 Mar 2010 22:04:08 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >>> If anything in the following is NOT correct according to >>> the emission theory, please state exactly what is wrong. >>> That is, answer question #1. >>> >>> 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. > >OK. >Since you haven't said otherwise, I will assume that you agree >that the following is a correct description 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. >>> >>> (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? >>> ============================================================ >> >> There are two 'wavelengths' in the source frame no matter how many differently >> moving observers look at it. > >Thank you for a clear answer. >So we agree: >The number of wavelengths in the wave is always two when t > T. >The path length of the wave front measured in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. > >Now we repeat the scenario with the source at the right end of >the rod, so the wave going in the other direction. > > > | | > |-----------* 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 left end >of the rod at x = X : > > > |** ** | > *--*--*--*--* phi = 4 pi > | ** **| > |-----|-----|-----------------------> x > 0 X L > > >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 X L > > > >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/2, so the path length is L/2 or >one wavelength. > >============================================================= >Now the question is: >How many wavelengths are there in the wave when t >= T? >Is it 2 or is it 1? >============================================================ > >The answer is obvious. When t > T, there will always be >2 wavelengths in the wave. > >Don't you agree? THERE ARE TWO WAVELENGTHS IN THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval. Both photons and photon density waves possess intrinsic spatial features that amount to an absolute wavelength, which does not vary with frame. You are confusing 'wavelength' with the distance the 'rest frame' moves relative to the source during the time taken for a pulse to travel a distance L/2. Put simply, you are using the light in the rod as a light clock to measure the speed of the rest frame relative to the rod. You are showing your lack of understanding of basic physics. Henry Wilson... ........provider of free physics lessons |