Sagnac Revisited. Why Paul, Tom and Jerry are Hopelessly Wrong.
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From: Paul B. Andersen on 10 Mar 2010 07:42 On 09.03.2010 22:53, Henry Wilson DSc wrote: > 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. Fine. So we agree. The number of wavelengths in the wave is always two whether the wave is going to the right or to the left. According to the emission theory the speed of light is c relative to the source, and the Galilean transform applies. So the motion of the rod in the stationary frame can't affect the number of wavelengths in the waves. > 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. Say, what are you talking about? :-) We agree that according to the emission theory, the number of wavelengths in the wave is always two whether the wave is going to the right or to the left. But the length of the trajectory of the wave front from emission to absorption (path length) for the right going wave is 3L/2 while the path length of the wave front of the left going wave is L/2, as measured in the stationary frame. Since one wavelength is L/2, these distances in the stationary frame are equal to three wavelengths and one wavelength respectively. But nobody would get the stupid idea that this means that there are any waves containing three or one wavelengths. Or would they? :-) ------------------------------------------ Now we modify the experiment a little. We bend the rod to a circle, and have one source emitting light in both directions. The light is guided (by mirrors) to move along the circular rod. The circular rod is rotating with the peripheral speed v as measured in the stationary frame. According to the emission theory, how many wavelengths are there now in the two contra moving waves? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 10 Mar 2010 15:58 On Wed, 10 Mar 2010 13:42:30 +0100, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 09.03.2010 22:53, Henry Wilson DSc wrote: >> On Tue, 09 Mar 2010 21:35:21 +0100, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >> >>>>> >>>>> S - - - - ->A >>>>> |-----------| -> v >>>>> L >>>>> |-----------------------------------> x >>>>> 0 >>>>> =================================== >>> >>> 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. > >Fine. >So we agree. >The number of wavelengths in the wave is always two whether >the wave is going to the right or to the left. > >According to the emission theory the speed of light is c >relative to the source, and the Galilean transform applies. >So the motion of the rod in the stationary frame can't >affect the number of wavelengths in the waves. The rod is inertial. Nothing in its frame can be affected by the relative movement of an observer. >> 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. > >Say, what are you talking about? :-) You are not talking about 'wavelength'. You are calculating distance moved by the other frame during one cycle. >We agree that according to the emission theory, the number >of wavelengths in the wave is always two whether the wave >is going to the right or to the left. The rod/source system is inertial. A 'wavelength' can be asssociated IN THE SOURCE FRAME with a generated EM wave emitted by the source . A INTRINSIC wavelength is associated with a single photon. BOTH lengths are absolute and frame invariant. >But the length of the trajectory of the wave front from >emission to absorption (path length) for the right going wave >is 3L/2 while the path length of the wave front of the left going >wave is L/2, as measured in the stationary frame. That's due to the doppler FREQUENCY shift. IT IS NOT a 'wavelength'. >Since one wavelength is L/2, these distances in the stationary >frame are equal to three wavelengths and one wavelength respectively. > >But nobody would get the stupid idea that this means that >there are any waves containing three or one wavelengths. > >Or would they? :-) Both YOU and Androcles would....because you are not talking about 'wavelength' at all. > >------------------------------------------ > >Now we modify the experiment a little. >We bend the rod to a circle, and have one source emitting >light in both directions. The light is guided (by mirrors) >to move along the circular rod. > >The circular rod is rotating with the peripheral speed v >as measured in the stationary frame. Don't use the word 'stationary'. As soon as you introduce rotation you have an entirely different problem. >According to the emission theory, how many wavelengths are >there now in the two contra moving waves? You must analyse it using the NonR frame. The path lengths differ by 2vt...or 2vt/L absolute wavelengths. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 12 Mar 2010 15:25 On 10.03.2010 21:58, Henry Wilson DSc wrote: > On Wed, 10 Mar 2010 13:42:30 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> Henry Wilson DSc wrote: >> >>> Paul B. Andersen wrote: >>>> 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. >> >> Fine. >> So we agree. >> The number of wavelengths in the wave is always two whether >> the wave is going to the right or to the left. >> >> According to the emission theory the speed of light is c >> relative to the source, and the Galilean transform applies. >> So the motion of the rod in the stationary frame can't >> affect the number of wavelengths in the waves. >> >> But the length of the trajectory of the wave front from >> emission to absorption (path length) for the right going wave >> is 3L/2 while the path length of the wave front of the left going >> wave is L/2, as measured in the stationary frame. > > That's due to the doppler FREQUENCY shift. :-) It's due to the motion of the rod. But let's summarize what we agree upon thus far. According to the emission theory, we have: Right going wave ------------------ Wave front start at x=0: | | *-----------| | | |-----------|-----------------------> x 0 L And reaches the end of the rod at x=X: | ** **| *--*--*--*--* |** ** | |-----------|-----|-----------------> x 0 L X The length of the trajectory of the wave front (path length) measured in the stationary frame is (X-0) = 3L/2. But when we are in steady state: "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." The absolute spatial length of the rod is two wavelengths. So the path length of the wave front in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. Left going wave --------------- The wave front starts at x=L | | |-----------* phi = 0 | | |-----------|-----------------------> x 0 L And reaches the end of the rod at x=X: |** ** | *--*--*--*--* phi = 4 pi | ** **| |-----|-----|-----------------------> x 0 X L The length of the trajectory of the wave front (path length) measured in the stationary frame is (L-X) = L/2. But when we are in steady state: "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." The absolute spatial length of the rod is two wavelengths. So the path length of the wave front in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. > IT IS NOT a 'wavelength'. WHAT is not a 'wavelength'? One 'wavelength' is L/2. It's a measure of length. You can obviously measure the length of anything in 'wavelengths'. It doesn't have to be a wave present to do that. So: >> Since one wavelength is L/2, these distances in the stationary >> frame are equal to three wavelengths and one wavelength respectively. >> >> But nobody would get the stupid idea that this means that >> there are any waves containing three or one wavelengths. >> >> Or would they? :-) > > Both YOU and Androcles would....because you are not talking about 'wavelength' > at all. But Ralph Rabbidge wouldn't? :-) >> ------------------------------------------ >> >> Now we modify the experiment a little. >> We bend the rod to a circle, and have one source emitting >> light in both directions. The light is guided (by mirrors) >> to move along the circular rod. >> >> The circular rod is rotating with the peripheral speed v >> as measured in the stationary frame. > > Don't use the word 'stationary'. > As soon as you introduce rotation you have an entirely different problem. Why this stupid quibbling? The 'stationary frame' is the screen. It is inertial. It is not rotating. It is a non-rotating frame. So: The circular rod is rotating with the peripheral speed v as measured in the stationary frame. The centre of the circle is stationary in the stationary frame. According to the emission theory we now have: On the circle in the stationary frame we can measure the position as an angle phi (in radians): x = 0 is at phi = 0 x = L/2 is at phi = pi x = L is at phi = 2 pi x = 3L/2 is at phi = 3 pi The clockwise going wave: ------------------------- Wave front start at phi = 0 And hits the absorber at phi = 3 pi The length of the trajectory of the wave front (path length) measured in the stationary frame is = 3L/2. But when we are in steady state: "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." The absolute spatial length of the rod is two wavelengths. So the path length of the wave front in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. The anti-clockwise going wave: ----------------------------- Wave front start at phi = 2 pi And hits the absorber at phi = pi The length of the trajectory of the wave front (path length) measured in the stationary frame is = L/2. But when we are in steady state: "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." The absolute spatial length of the rod is two wavelengths. So the path length of the wave front in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. >> According to the emission theory, how many wavelengths are >> there now in the two contra moving waves? BEHOLD, BEHOLD: > You must analyse it using the NonR frame. The path lengths differ by 2vt...or > 2vt/L absolute wavelengths. ######################################################################## # Since one wavelength is L/2, these path lengths in the stationary # frame are equal to three wavelengths and one wavelength respectively. # # But nobody would get the stupid idea that this means that # there are any waves containing three or one wavelengths. # # Or would they? :-) ########################################################################## You just did! :-) Try again. The question was: According to the emission theory, how many wavelengths are there now in the two contra moving waves? -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 12 Mar 2010 17:20 On Fri, 12 Mar 2010 21:25:47 +0100, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 10.03.2010 21:58, Henry Wilson DSc wrote: >> On Wed, 10 Mar 2010 13:42:30 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>> Henry Wilson DSc wrote: >>> >>>> Paul B. Andersen wrote: >>>>> 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. >>> >>> Fine. >>> So we agree. >>> The number of wavelengths in the wave is always two whether >>> the wave is going to the right or to the left. >>> >>> According to the emission theory the speed of light is c >>> relative to the source, and the Galilean transform applies. >>> So the motion of the rod in the stationary frame can't >>> affect the number of wavelengths in the waves. >>> >>> But the length of the trajectory of the wave front from >>> emission to absorption (path length) for the right going wave >>> is 3L/2 while the path length of the wave front of the left going >>> wave is L/2, as measured in the stationary frame. >> >> That's due to the doppler FREQUENCY shift. > >:-) It's due to the motion of the rod. > >But let's summarize what we agree upon thus far. > >According to the emission theory, we have: > >Right going wave >------------------ >Wave front start at x=0: > > | | > *-----------| > | | > |-----------|-----------------------> x > 0 L > >And reaches the end of the rod at x=X: > > | ** **| > *--*--*--*--* > |** ** | > |-----------|-----|-----------------> x > 0 L X > >The length of the trajectory of the wave front (path length) >measured in the stationary frame is (X-0) = 3L/2. > >But when we are in steady state: > "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD > no matter who looks at it. > The distance between crests in the source frame defines > an absolute spatial interval." > >The absolute spatial length of the rod is two wavelengths. >So the path length of the wave front in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. It's utterly irrelvant to anything. >Left going wave >--------------- > >The wave front starts at x=L > | | > |-----------* phi = 0 > | | > |-----------|-----------------------> x > 0 L > >And reaches the end of the rod at x=X: > > |** ** | > *--*--*--*--* phi = 4 pi > | ** **| > |-----|-----|-----------------------> x > 0 X L > >The length of the trajectory of the wave front (path length) >measured in the stationary frame is (L-X) = L/2. > >But when we are in steady state: > "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD > no matter who looks at it. > The distance between crests in the source frame defines > an absolute spatial interval." > >The absolute spatial length of the rod is two wavelengths. >So the path length of the wave front in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. It's utterly irrelevant to anything. >> IT IS NOT a 'wavelength'. > >WHAT is not a 'wavelength'? The distance you 'rest frame' moves during one cycle of the 'wave'. What wave are you actually talking about anyway? >One 'wavelength' is L/2. It's a measure of length. >You can obviously measure the length of anything in 'wavelengths'. >It doesn't have to be a wave present to do that. > >So: >>> Since one wavelength is L/2, these distances in the stationary >>> frame are equal to three wavelengths and one wavelength respectively. >>> Now we modify the experiment a little. >>> We bend the rod to a circle, and have one source emitting >>> light in both directions. The light is guided (by mirrors) >>> to move along the circular rod. >>> >>> The circular rod is rotating with the peripheral speed v >>> as measured in the stationary frame. >> >> Don't use the word 'stationary'. >> As soon as you introduce rotation you have an entirely different problem. > >Why this stupid quibbling? > >The 'stationary frame' is the screen. >It is inertial. It is not rotating. >It is a non-rotating frame. > >So: >The circular rod is rotating with the peripheral speed v >as measured in the stationary frame. >The centre of the circle is stationary in the stationary frame. > >According to the emission theory we now have: > >On the circle in the stationary frame we can measure >the position as an angle phi (in radians): >x = 0 is at phi = 0 >x = L/2 is at phi = pi >x = L is at phi = 2 pi >x = 3L/2 is at phi = 3 pi > >The clockwise going wave: >------------------------- >Wave front start at phi = 0 >And hits the absorber at phi = 3 pi > >The length of the trajectory of the wave front (path length) >measured in the stationary frame is = 3L/2. > >But when we are in steady state: > "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD > no matter who looks at it. > The distance between crests in the source frame defines > an absolute spatial interval." > >The absolute spatial length of the rod is two wavelengths. >So the path length of the wave front in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. there is no 'path length' in the inertial situation. Path lengths only exist in the rotating experiment. >The anti-clockwise going wave: >----------------------------- >Wave front start at phi = 2 pi >And hits the absorber at phi = pi > >The length of the trajectory of the wave front (path length) >measured in the stationary frame is = L/2. > >But when we are in steady state: > "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD > no matter who looks at it. > The distance between crests in the source frame defines > an absolute spatial interval." > >The absolute spatial length of the rod is two wavelengths. >So the path length of the wave front in the stationary frame >is utterly irrelevant to the number of wavelengths in the wave. Where would you get that idea? The 'wavelength' is an bsolute distance. If two paths are of different length, they will contain different numbers of wavelengths. >>> According to the emission theory, how many wavelengths are >>> there now in the two contra moving waves? > >BEHOLD, BEHOLD: > >> You must analyse it using the NonR frame. The path lengths differ by 2vt...or >> 2vt/L absolute wavelengths. > >######################################################################## ># Since one wavelength is L/2, these path lengths in the stationary ># frame are equal to three wavelengths and one wavelength respectively. ># ># But nobody would get the stupid idea that this means that ># there are any waves containing three or one wavelengths. ># ># Or would they? :-) >########################################################################## > >You just did! :-) > >Try again. > >The question was: >According to the emission theory, how many wavelengths are >there now in the two contra moving waves? You are raving about inertial movement when it is rotation around a closed ring that is important. I cannot see why you are having so much trouble understanding this. It is very simple. Henry Wilson... ........provider of free physics lessons
From: Paul B. Andersen on 15 Mar 2010 06:44
On 12.03.2010 23:20, Henry Wilson DSc wrote: > On Fri, 12 Mar 2010 21:25:47 +0100, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: > >> On 10.03.2010 21:58, Henry Wilson DSc wrote: >>> On Wed, 10 Mar 2010 13:42:30 +0100, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> >>>> Henry Wilson DSc wrote: >>>> >>>>> Paul B. Andersen wrote: >>>>>> 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. >>>> >>>> Fine. >>>> So we agree. >>>> The number of wavelengths in the wave is always two whether >>>> the wave is going to the right or to the left. >>>> >>>> According to the emission theory the speed of light is c >>>> relative to the source, and the Galilean transform applies. >>>> So the motion of the rod in the stationary frame can't >>>> affect the number of wavelengths in the waves. >>>> >>>> But the length of the trajectory of the wave front from >>>> emission to absorption (path length) for the right going wave >>>> is 3L/2 while the path length of the wave front of the left going >>>> wave is L/2, as measured in the stationary frame. >>> >>> That's due to the doppler FREQUENCY shift. >> >> :-) It's due to the motion of the rod. >> >> But let's summarize what we agree upon thus far. >> >> According to the emission theory, we have: >> >> Right going wave >> ------------------ >> Wave front start at x=0: >> >> | | >> *-----------| >> | | >> |-----------|-----------------------> x >> 0 L >> >> And reaches the end of the rod at x=X: >> >> | ** **| >> *--*--*--*--* >> |** ** | >> |-----------|-----|-----------------> x >> 0 L X >> >> The length of the trajectory of the wave front (path length) >> measured in the stationary frame is (X-0) = 3L/2. >> >> But when we are in steady state: >> "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >> no matter who looks at it. >> The distance between crests in the source frame defines >> an absolute spatial interval." >> >> The absolute spatial length of the rod is two wavelengths. >> So the path length of the wave front in the stationary frame >> is utterly irrelevant to the number of wavelengths in the wave. > > It's utterly irrelvant to anything. > >> Left going wave >> --------------- >> >> The wave front starts at x=L >> | | >> |-----------* phi = 0 >> | | >> |-----------|-----------------------> x >> 0 L >> >> And reaches the end of the rod at x=X: >> >> |** ** | >> *--*--*--*--* phi = 4 pi >> | ** **| >> |-----|-----|-----------------------> x >> 0 X L >> >> The length of the trajectory of the wave front (path length) >> measured in the stationary frame is (L-X) = L/2. >> >> But when we are in steady state: >> "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >> no matter who looks at it. >> The distance between crests in the source frame defines >> an absolute spatial interval." >> >> The absolute spatial length of the rod is two wavelengths. >> So the path length of the wave front in the stationary frame >> is utterly irrelevant to the number of wavelengths in the wave. > > It's utterly irrelevant to anything. > >>> IT IS NOT a 'wavelength'. >> >> WHAT is not a 'wavelength'? > > The distance you 'rest frame' moves during one cycle of the 'wave'. > What wave are you actually talking about anyway? > >> One 'wavelength' is L/2. It's a measure of length. >> You can obviously measure the length of anything in 'wavelengths'. >> It doesn't have to be a wave present to do that. >> >> So: >>>> Since one wavelength is L/2, these distances in the stationary >>>> frame are equal to three wavelengths and one wavelength respectively. > >>>> Now we modify the experiment a little. >>>> We bend the rod to a circle, and have one source emitting >>>> light in both directions. The light is guided (by mirrors) >>>> to move along the circular rod. >>>> >>>> The circular rod is rotating with the peripheral speed v >>>> as measured in the stationary frame. >>> >>> Don't use the word 'stationary'. >>> As soon as you introduce rotation you have an entirely different problem. >> >> Why this stupid quibbling? >> >> The 'stationary frame' is the screen. >> It is inertial. It is not rotating. >> It is a non-rotating frame. >> >> So: >> The circular rod is rotating with the peripheral speed v >> as measured in the stationary frame. >> The centre of the circle is stationary in the stationary frame. >> >> According to the emission theory we now have: >> >> On the circle in the stationary frame we can measure >> the position as an angle phi (in radians): >> x = 0 is at phi = 0 >> x = L/2 is at phi = pi >> x = L is at phi = 2 pi >> x = 3L/2 is at phi = 3 pi >> >> The clockwise going wave: >> ------------------------- >> Wave front start at phi = 0 >> And hits the absorber at phi = 3 pi >> >> The length of the trajectory of the wave front (path length) >> measured in the stationary frame is = 3L/2. >> >> But when we are in steady state: >> "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >> no matter who looks at it. >> The distance between crests in the source frame defines >> an absolute spatial interval." >> >> The absolute spatial length of the rod is two wavelengths. >> So the path length of the wave front in the stationary frame >> is utterly irrelevant to the number of wavelengths in the wave. > > there is no 'path length' in the inertial situation. > > Path lengths only exist in the rotating experiment. Please don't state stupidietes like this. Wave front start at x=0: | | *-----------| | | |-----------|-----------------------> x 0 L And reaches the end of the rod at x=X: | ** **| *--*--*--*--* |** ** | |-----------|-----|-----------------> x 0 L X The 'path length' is the length of the trajectory of the wave front measured in the stationary frame. In this case it is (X-0) = 3L/2. > >> The anti-clockwise going wave: >> ----------------------------- >> Wave front start at phi = 2 pi >> And hits the absorber at phi = pi >> >> The length of the trajectory of the wave front (path length) >> measured in the stationary frame is = L/2. >> >> But when we are in steady state: >> "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD >> no matter who looks at it. >> The distance between crests in the source frame defines >> an absolute spatial interval." >> >> The absolute spatial length of the rod is two wavelengths. >> So the path length of the wave front in the stationary frame >> is utterly irrelevant to the number of wavelengths in the wave. > > Where would you get that idea? > The 'wavelength' is an bsolute distance. > If two paths are of different length, they will contain different numbers of > wavelengths. Indeed. The path lengths measured in wavelengths are different. Don't you read what I write? ######################################################################## # Since one wavelength is L/2, these path lengths in the stationary # frame are equal to three wavelengths and one wavelength respectively. ####################################################################### But as YOU so correctly put it: "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." So the path length of the wave front in the stationary frame is utterly irrelevant to the number of wavelengths in the wave. So: ######################################################################## # But nobody would get the stupid idea that this means that # there are any waves containing three or one wavelengths. # # Or would they? :-) ########################################################################## >>>> According to the emission theory, how many wavelengths are >>>> there now in the two contra moving waves? >> >> BEHOLD, BEHOLD: >> >>> You must analyse it using the NonR frame. The path lengths differ by 2vt...or >>> 2vt/L absolute wavelengths. >> >> ######################################################################## >> # Since one wavelength is L/2, these path lengths in the stationary >> # frame are equal to three wavelengths and one wavelength respectively. >> # >> # But nobody would get the stupid idea that this means that >> # there are any waves containing three or one wavelengths. >> # >> # Or would they? :-) >> ########################################################################## >> >> You just did! :-) >> >> Try again. >> >> The question was: >> According to the emission theory, how many wavelengths are >> there now in the two contra moving waves? > > You are raving about inertial movement when it is rotation around a closed ring > that is important. > I cannot see why you are having so much trouble understanding this. It is very > simple. Quite. It is very simple. The length of the rod is L, which according to the emission theory is 'an absolute spatial interval'. The absolute spatial interval 'one wavelength' is L/2. So "THERE ARE TWO WAVELENGTHS IN [the wave on] THE ROD no matter who looks at it. The distance between crests in the source frame defines an absolute spatial interval." I cannot see why you are having so much trouble understanding this. It is very simple. So maybe you now can answer the simple question? According to the emission theory, how many wavelengths are there now in the two contra moving waves? -- Paul http://home.c2i.net/pb_andersen/ |