Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 30 Nov 2005 15:08 "Henri Wilson" <HW@..> wrote in message news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... > On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:pjumo1p9d978pli6l3llvkvaq5edjaprrr(a)4ax.com... >>> On Mon, 28 Nov 2005 20:07:21 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>>>"Henri Wilson" <HW@..> wrote in message >>>>news:06c7o1ptp360adjulvbpol9ved8io57jk8(a)4ax.com... >>... >>>>> Maybe we should start all over again. >>>> >>>>I'll see if I can start from the beginning then, >>>>maybe some aspect is slipping past without you >>>>noticing. I am still just trying to get you to >>>>see the consequences of your own drawing though, >>>>I'm not introducing anything new. >>>> >>>> http://www.users.bigpond.com/hewn/sagnac.jpg >>>> >>>>Path length at rest is distance L from point A to >>>>point C in time t. The light moves at speed c so >>>> >>>> t = L/c. >>>> >>>>When the table is rotating at constant speed, light >>>>is emitted at point A but the table moves a distance >>>>vt from point C to point D while the light is in >>>>transit. I'll add a single quote to indicate the >>>>values for this case where they differ from the >>>>static situation. >>>> >>>>The speed of the light in the direction you have >>>>shown has been increased by Ritz to >>>> >>>> c' = c + v/sqrt(2) >>>> >>>>where v is the tangential speed of the source. The >>>>path length increases to >>>> >>>> L' = L + vt'/sqrt(2) >>>> >>>>and of course >>>> >>>> t' = L'/c' >>>> >>>>Remember we are interested in the difference between >>>>the two paths and the other has been shortened but >>>>the light slowed. The results are symmetrical so I'll >>>>only do the maths for one. (Just use -v instead of v >>>>for the other.) >>>> >>>>To save typing, note that v appears with sqrt(2) in >>>>both so let >>>> >>>> V = v/sqrt(2) >>>> >>>>so: >>>> >>>> c' = c + V >>>> >>>> L' = L + Vt' >>>> >>>>Solve for t': >>>> >>>> t' = L'/c' >>>> >>>> c't' = L' >>>> >>>>Substituting c' and L' we get: >>>> >>>> (c + V)t' = (L + Vt') >>>> >>>> ct' + Vt' = L + Vt' >>>> >>>> ct' = L >>>> >>>> t' = L/c >>>> >>>> t' = t >>>> >>>>So constant speed rotation produces no change to the >>>>time taken for the light to travel the increased path >>>>length, the increased speed compensates. That means >>>>no fringe displacement because the displacement is >>>>determined by the time difference between the forward >>>>and backward paths. They are both equal (at t) when >>>>the table is stationary so the are still equal when >>>>it is rotating at constant speed. >>>> >>>>Notice also that the sqrt(2) factor affects both the >>>>path length and the boost in speed through the geometry >>>>so a different number of mirrors would give a different >>>>factor but it would still cancel out. >>> >>> Yes I have now done these calculations myself and I think you are right, >>> if we >>> ignore second order effects and make certain assumptions about the way >>> light >>> reflects from each mirror. >>> >>>>Now consider the case with acceleration. Definitions >>>>are critical here because v is now a function of time, >>>>it is increasing linearly. I'll add double quotes for >>>>this case. >>>> >>>>At some instant, a wavefront is emitted from the >>>>source. At that time, the tangential speed of the >>>>source is v. The trick here is that the light keeps >>>>the same speed once it is emitted while the table >>>>is accelerating hence as before >>>> >>>> c" = c + V >>>> >>>>so >>>> >>>> c" = c' >>>> >>>>When the light is in transit for time t", the table >>>>speeds up by a factor of >>>> >>>> dv = a t" >>>> >>>>The average increase during t" is half that so the >>>>distance by which the table moves more than the >>>>constant speed case is >>>> >>>> dL = (a t" / 2) t" >>>> >>>>The path length shown on the diagram is given by >>>> >>>> L" = L + Vt" + a t"^2 / 2 >>> >>> (How about L" = L + Vt" + A t"^2 / 2, where A=a/sqrt2) >> >>Yes, you are correct. It reduces the amplitude >>of the signal you would get according to Ritz >>but doesn't change the logic, the predicted >>result is still proportional to acceleration, >>not speed. >> >>>>Obviously >>>> >>>> t" = L"/c" >>> >>> (To first order only. L is not constant... but the 2nd order effect is >>> very >>> small.) >> >>No, it is exact. L" and c" are specific values for >>any particular wavefront so the time taken is the >>distance divided by the speed. >> >>The values of L" and c" for the the next wavefront >>will both be greater but these equations hold for >>each individually. It is not until later that it >>is shown that t" = t and then we can conclude that >>t" is the same for all wavefronts. At this stage >>it might have a different value in each case. >> >>>> t" = (L' + dL) / c' >>>> >>>> t" = L'/c' + dL/c' >>>> >>>> t" = t + dt >>>> >>>>where >>>> >>>> dt = (a t"^2 / 2) / c' >>>> >>>>That is what you should expect when you realise that, >>>>if a=0, we are back to the previous situation and the >>>>Ritzian change to the speed of the light cancels the >>>>effect of the Vt" term as it did before. >>> >>> OK so far. (apart from the a/sqrt2) >> >>Excellent, we agree entirely. >> >>>>For the other path however, the acceleration _decreases_ >>>>the path length so this time we do not get cancellation >>>>and there is going to be a fringe displacement that >>>>depends on the acceleration. >>> >>> I don't think that is right George. The V terms still cancel but the A >>> terms >>> for both directions still add...so we get double the effect. >>> >>> Maybe I'm missing something here. >> >>You misread me somehow, what I said was the same, >>there IS going to be a fringe displacement. Maybe >>I wasn't clear enough but we are in complete >>agreement. <snip my bit on "fringe displacement"> >>>> >>>>For this speed profile: >>>> >>>> _________ >>>> / \ >>>> / \ ^ >>>> __________/ \_________ | speed >>>> | >>>> __________________________________ >>>> >>>> ------> >>>> time >>>> >>>> Ritz predicts this time difference between the beams: >>>> >>>> __ >>>> +ve | | ^ >>>> | | | time difference >>>> | | | >>>> 0 _______| |_______ ________ >>>> | | >>>> ------> | | >>>> time | | >>>> -ve |__| >>> >>> That's all very well George. ..but quite irrelevant in light of my >>> latest >>> findings. >> >>Perhaps, but it's taken months to get to the point >>where we agree it so I want to make sure it is noted. >>I don't want to have to come back and do this all >>again next year. > > I will have a good look at this before I agree 100%. OK, I had gone a little farther than the numbers above warranted. I have deleted the translation to fringe displacement since that depends on understanding the interference aspects which we are still discussing below. The difference between the time taken to travel from source to detector over the two paths is what is given by the equations above so you should find that easier to agree, the derivation is all there. > There seems to be something logically wrong here. Intuitively, think of grandpa and the kids on the carousel. If they run round it at equal speed relative to the surface, they will get back to him at the same time regardless of the speed of rotation. If you want to analyse the accelerated case, put the kids on roller skates and have grandma on the ground give the carousel an extra push after the kids are on their way. The skates mean their speed relative to the ground doesn't change once they have started, but grandpa's speed changes so he meets one child before the other. >>When the table is turning at constant speed, it is >>light emitted in the A-D direction that hits the >>detector at D instead of light emitted in the A-C >>direction when the table is stationary. The >>consequence is the vt term which we have agreed >>above is cancelled by the effect of the boost in >>speed from c to c'. > > It is for path from the source to the first mirror anyway. I'm not > completely > convinced about what happens to the subsequent sections. Point A can be the source and C/D can be the first mirror, or A can be the first mirror and C/D the second and so on, the analysis applies to each leg. Obviously the total times are just multiplied by four. I'll stop there in this reply and see if we can agree up to here. I have some stuff on interferometers that I'll post later tonight, time permitting. I'm out every night from tomorrow to Saturday so might not get a chance to reply again until Sunday. George
From: Henri Wilson on 30 Nov 2005 15:37 On Wed, 30 Nov 2005 20:08:22 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... >> On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >>>> >>>> Maybe I'm missing something here. >>> >>>You misread me somehow, what I said was the same, >>>there IS going to be a fringe displacement. Maybe >>>I wasn't clear enough but we are in complete >>>agreement. > ><snip my bit on "fringe displacement"> > >>>>> >>>>>For this speed profile: >>>>> >>>>> _________ >>>>> / \ >>>>> / \ ^ >>>>> __________/ \_________ | speed >>>>> | >>>>> __________________________________ >>>>> >>>>> ------> >>>>> time >>>>> >>>>> > >Ritz predicts this time difference between the beams: > >>>>> >>>>> __ >>>>> +ve | | ^ >>>>> | | | time difference >>>>> | | | >>>>> 0 _______| |_______ ________ >>>>> | | >>>>> ------> | | >>>>> time | | >>>>> -ve |__| >>>> >>>> That's all very well George. ..but quite irrelevant in light of my >>>> latest >>>> findings. >>> >>>Perhaps, but it's taken months to get to the point >>>where we agree it so I want to make sure it is noted. >>>I don't want to have to come back and do this all >>>again next year. >> >> I will have a good look at this before I agree 100%. > >OK, I had gone a little farther than the numbers above >warranted. I have deleted the translation to fringe >displacement since that depends on understanding the >interference aspects which we are still discussing >below. The difference between the time taken to travel >from source to detector over the two paths is what is >given by the equations above so you should find that >easier to agree, the derivation is all there. I'm not certain about what happens at each reflection. > >> There seems to be something logically wrong here. > >Intuitively, think of grandpa and the kids on the >carousel. If they run round it at equal speed relative >to the surface, they will get back to him at the same >time regardless of the speed of rotation. Ah! But there is a difference. Let's say the kids run around at the same speed as the carousel. One kid remains in the one spot. He feels no centrifugal force. The other has four times the C.F. Have you considered that light might experience something similar? In which case, the two beams will experience some kind of effect that is related to ABSOLUTE rotation. > >If you want to analyse the accelerated case, put the >kids on roller skates and have grandma on the ground >give the carousel an extra push after the kids are on >their way. The skates mean their speed relative to the >ground doesn't change once they have started, but >grandpa's speed changes so he meets one child before >the other. Yes. OK. >>>When the table is turning at constant speed, it is >>>light emitted in the A-D direction that hits the >>>detector at D instead of light emitted in the A-C >>>direction when the table is stationary. The >>>consequence is the vt term which we have agreed >>>above is cancelled by the effect of the boost in >>>speed from c to c'. >> >> It is for path from the source to the first mirror anyway. I'm not >> completely >> convinced about what happens to the subsequent sections. > >Point A can be the source and C/D can be the first >mirror, or A can be the first mirror and C/D the >second and so on, the analysis applies to each leg. >Obviously the total times are just multiplied by four. What happens at each reflection is far from obvious George. ...but for the moment I will agree with you.. > >I'll stop there in this reply and see if we can agree >up to here. I have some stuff on interferometers that >I'll post later tonight, time permitting. I'm out >every night from tomorrow to Saturday so might not get >a chance to reply again until Sunday. We have a lot to think about. The 'centrifugal force' idea for a start. > >George > HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong".
From: George Dishman on 30 Nov 2005 17:34 "Henri Wilson" <HW@..> wrote in message news:bq2so15sdi9nvu4au9cjcsmogjft4nedo0(a)4ax.com... > On Wed, 30 Nov 2005 20:08:22 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... >>> On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> > > >>>>> >>>>> Maybe I'm missing something here. >>>> >>>>You misread me somehow, what I said was the same, >>>>there IS going to be a fringe displacement. Maybe >>>>I wasn't clear enough but we are in complete >>>>agreement. >> >><snip my bit on "fringe displacement"> >> >>>>>> >>>>>>For this speed profile: >>>>>> >>>>>> _________ >>>>>> / \ >>>>>> / \ ^ >>>>>> __________/ \_________ | speed >>>>>> | >>>>>> __________________________________ >>>>>> >>>>>> ------> >>>>>> time >>>>>> >>>>>> >> >>Ritz predicts this time difference between the beams: >> >>>>>> >>>>>> __ >>>>>> +ve | | ^ >>>>>> | | | time difference >>>>>> | | | >>>>>> 0 _______| |_______ ________ >>>>>> | | >>>>>> ------> | | >>>>>> time | | >>>>>> -ve |__| >>>>> >>>>> That's all very well George. ..but quite irrelevant in light of my >>>>> latest >>>>> findings. >>>> >>>>Perhaps, but it's taken months to get to the point >>>>where we agree it so I want to make sure it is noted. >>>>I don't want to have to come back and do this all >>>>again next year. >>> >>> I will have a good look at this before I agree 100%. >> >>OK, I had gone a little farther than the numbers above >>warranted. I have deleted the translation to fringe >>displacement since that depends on understanding the >>interference aspects which we are still discussing >>below. The difference between the time taken to travel >>from source to detector over the two paths is what is >>given by the equations above so you should find that >>easier to agree, the derivation is all there. > > I'm not certain about what happens at each reflection. Obviously that depends on your understanding of Ritzian theory. I can se two possibilities, one being that the speed of the reflected light is the same as that of the incident light and the other being that it is absorbed and re-emitted at c relative to the mirror. Luckily in our case, both interpretations give the same result because the relative speed of the incident light is c. If you come up with a third interpretation then you would have to work out the conequences. >>> There seems to be something logically wrong here. >> >>Intuitively, think of grandpa and the kids on the >>carousel. If they run round it at equal speed relative >>to the surface, they will get back to him at the same >>time regardless of the speed of rotation. > > Ah! But there is a difference. > > Let's say the kids run around at the same speed as the carousel. > One kid remains in the one spot. He feels no centrifugal force. > The other has four times the C.F. > > Have you considered that light might experience something similar? Well not specifically but yes there is a similar effect, it is just "radiation pressure". The light will push the mirrors outwards very slightly. In the case of the discrete mirrors we examine above, the light travels in straight lines so there is no effect between reflections. Anyway supppose the radiation pressure caused some tiny outward movement of the mirrors (or expansion of the fibre loop in an iFOG), the increased path length still affects both beams equally. That's the beauty of the experiment, both beams traverse the same path in opposite directions so most effects like this cancel. In reality, thermal expansion will be far larger than radiation pressure effects but the principle still applies. > In which case, the two beams will experience some kind of effect that is > related to ABSOLUTE rotation. Exactly, they experience the same effect but you are looking for opposite effects on the beams. >>If you want to analyse the accelerated case, put the >>kids on roller skates and have grandma on the ground >>give the carousel an extra push after the kids are on >>their way. The skates mean their speed relative to the >>ground doesn't change once they have started, but >>grandpa's speed changes so he meets one child before >>the other. > > Yes. OK. > >>>>When the table is turning at constant speed, it is >>>>light emitted in the A-D direction that hits the >>>>detector at D instead of light emitted in the A-C >>>>direction when the table is stationary. The >>>>consequence is the vt term which we have agreed >>>>above is cancelled by the effect of the boost in >>>>speed from c to c'. >>> >>> It is for path from the source to the first mirror anyway. I'm not >>> completely >>> convinced about what happens to the subsequent sections. >> >>Point A can be the source and C/D can be the first >>mirror, or A can be the first mirror and C/D the >>second and so on, the analysis applies to each leg. >>Obviously the total times are just multiplied by four. > > What happens at each reflection is far from obvious George. For the two obvious interpretations, the rules of reflection are just what we are used to, angle of reflection equals angle of incidence since the speeds are both c relative to the mirror. > ..but for the moment I will agree with you.. That's good. It's based on sound physics but please do continue to check to see if there is any part you doubt, I'm happy to come back to it if you think you have found a problem. >>I'll stop there in this reply and see if we can agree >>up to here. I have some stuff on interferometers that >>I'll post later tonight, time permitting. I'm out >>every night from tomorrow to Saturday so might not get >>a chance to reply again until Sunday. > > We have a lot to think about. > The 'centrifugal force' idea for a start. Nah, that's trivial. Interference is the next tricky bit. George
From: George Dishman on 30 Nov 2005 18:39 "Henri Wilson" <HW@..> wrote in message news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... > On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:pjumo1p9d978pli6l3llvkvaq5edjaprrr(a)4ax.com... >>> On Mon, 28 Nov 2005 20:07:21 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>>>"Henri Wilson" <HW@..> wrote in message >>>>news:06c7o1ptp360adjulvbpol9ved8io57jk8(a)4ax.com... <continued> >>Having arrived at the point where we share a view on >>the path length analysis, I'm happy to move on to look >>at the next point you raise. >> >>> We have to go right back to basics about why fringes are formed at all. >>> >>> You know how to make 'line fringes' rather than circular ones using an >>> 'optical wedge'.. ..so let's use line fringes. >> >>They aren't really the same Henri, lines are produced >>by a lateral displacement for sources at equal path >>length whereas in the Sagnac the circular fringes are >>produced by sources in line at different distances >>from the screen. > > I have never seen the fringes in a sagnac interferometer. are you certain > that > circular fringes are used. Yes. There's a photo of the output from a Michelson interferometer without the 'wedge' offset half way down this page: http://www.3dimagery.com/michelsn.html They use the circular pattern to align the system, any angle between the paths moves the centre of the circle off the screen. In the Sagnac configuration, both beams reflect around _all_ the mirrors so any angle applied to one mirror affects both beams, hence you are going to get a circular pattern. The page below explains in more detail. > It doesn't matter all that much because whatever method is used to measure > fringe displacement will focus on a fairly straight section anyway. Well as I have mentioned a few times, in an iFOG the fringes aren't measured at all, but in the lab experiment they can be. You just measure the radius of the fringes so as you say it isn't too different. However the fact that the fringes are circular tells you a lot about the geometry that is responsible for the effect. >>> If the beams are perfectly parallel and coherent, they will >>> alternatively >>> reinforce or destructively interfere when they unite at the final >>> surface >>> and >>> lines will appear. >> >>For line fringes, if they are perfectly parallel, you >>get no fringes, the signals maintain constant phase >>across the screen. > > That's right. That's why in the MMX, one mirror can be slightly tilted to > make > an optical wedge. In that case, the beams meet with sinusoidally varying > phase > differences across the final screen. > Frankly, I'm not quite sure how circular fringes are formed in either > interferometer. OK, this page explains it though it isn't particularly well written: http://www.phy.davidson.edu/StuHome/cabell_f/diffractionfinal/pages/Michelson.htm One way of thinking of it is that, if the plates are exactly parallel, you get circles. If they are at an angle, the centre of the circles is pushed far off to the side and you basically see arcs which are very close to straight lines. Again there are photos near the bottom. > If the beams were both perfectly parallel and coherent, there > should be no phase difference at any point across the whole viewing area. > I suppose if one considers that the two uniting wavefronts are spherical > rather > than flat, one gets a type 'Newton's rings' pattern...but why aren't the > wavefronts flat? The formation is very similar to Newton's rings the fringes are at such an angle that the distance between the images of the mirror is a multiple of the wavelength when measured along the slope. >>I think it would be better to stick >>with the actual Sagnac configuration where you do get >>fringes if the beams are parallel but from different >>distances. I'll maybe sketch that if I get some time >>later in the week. > > If you are going to use spherical wavefronts with different diameters, I > already don't like it. > I was under the impression that the two sagnac paths were equal. When it is stationary, they are. They must be because the paths are actually the same, the light just goes round in opposite drections. The change in path length comes from the effects of motion - back to the diagram: http://www.briar.demon.co.uk/Henri/sagnac.gif One path is increased while the other is decreased though according to Ritz it should be only acceleration that causes a change, not rotation at constant speed. Think of the images of the source seen through the two paths. At rest we get one source illuminating a screen: <-- L --> | S + | source screen L is the total path length, four times that in the diagram of the single leg. When the system is still, you get uniform phase across the whole screen so no fringes. When motion causes a change in the lengths of dL (again four times the value in the diagram of the single leg) we get interferenece: <-- L --> | S | S' + | <->|<-> dL dL In the lab experiment you see circular fringes on the screen. In an iFOG, the intensity at the "+" sign is measured and depends on the phase. As we discussed before, it is actually the ampltude of the modulation that is used so it's an AC signal which can be amplified with accurately known gain hence a small fraction of a fring shift can be measured but that's an engineering detail. This Java applet let's you play with the effect although it's a bit tricky at low values of separation. http://www.physics.uq.edu.au/people/mcintyre/applets/michelson/michel.html It also illustrates the difference for an iFOG. They don't count fringes, they have a photodiode measuring the intensity at the centre. Think of it as being just the pixel at the centre of the white cross. A graph of the value is shown to the right of the simulated view. <snip bit dealt with in earlier reply> >>The other aspect is a possible change in the angle >>between the beams. To sort that out, start by >>thinking about the car, the duck and the goose: >> >>> .. I like your >>> analogy for the four mirror case so let's look in a >>> bit more detail. First add a goose following the car >>> and another hunter shooting back at it: >>> >>> >>> <- Car >>> * >>> / \guns >>> / | \ >>> / | \ >>> / | \ >>> / 44 | 46 \ >>> / | \ >>> | / | \ ^ >>> v * + * | >>> Duck Goose >>> >>> >>> The "+" in the middle indicates the centre of the >>> roundabout and all are at equal distance R form >>> that and moving at the same speed, V, in the >>> directions indicated by the arrows. >>> >>> Now add a large sheet of paper stretched between >>> the duck, the car and the goose. The guns are fired >>> simultaneously and the lead bullets just skim the >>> surface of the paper leaving a drawn line. >> >>When nothing is moving, the guns shoot at 45 degrees >>to a radial line from the centre. >> >>When all are moving anti-clockwise as shown, the >>leading gun must aim ahead of the duck so shoots >>at less than 45 degrees (shown as "44") while the >>other gun must shoot at more than 45 degrees, shown >>as 46. >> >>The key here is that the angle between the guns >>remains 90 degrees because one increases while the >>other decreases. > > There is a bit of a trap here. > You have to consider the velocity of the gun as the bullet is fired. That is aberration. > The two angles are not quite symetrical about 45 deg. No, it decreases the 44 figure and increases the 46 figure so the total is still 90 degrees. > In my diagram, I have represented those components by F-D and G-E. > If you draw them on your diagram in the two directions, you will see what > I > mean. I removed them in my copy because you are counting the effect twice. The diagram is drawn in the lab frame so the motion of the source affects the speed of the light at launch. The light is emitted at point A. The source is moving towrds B but once launched that doesn't matter. In the lab frame the target is moving vertically downwards from C towrds D, not diagonally from C to G. (The latter would be the case in the source frame.) > So the angle does not remain at exactly 90. If you think about it, you will see it does. Bottom line we agreed was that the light must hit the detector to have an effect. The paths needed for that to happen are shown on the Java I did some time ago http://www.briar.demon.co.uk/Henri/SagnacAngles.html >>Now look at your animation >> >> www.users.bigpond.com/hewn/sagnac1.exe >> >>When it finishes, the enlargement shows that both >>beams must move slightly anti-clockwise, and by the >>same amount, to hit the detector. That means that >>angle between them will stay the same. >> >>Once re-combined by the beam splitter, if they >>started parallel when the table isn't turning then >>they remain parallel when it is. The fringes must >>be purely a result of the changed path lengths and >>Ritz cannot explain that. >> >>Remember also that in the iFOG case, we are measuring >>the actual time delay of the modulating waveform on >>the signal. There is no fringe pattern produced, the >>sensor only ever measures the intensity at the centre. >> >>George > > George, I'll try to draw what I'm trying to say. Use fixed pitch fonts. > > Consider two beams meeting each other as in a sagnac. They alternatively > reinforce and destruct across the screen where they meet. The +'s mark the > points where the beams reinforce and bright fringes appear. (It could be a > section of a circular fringe pattern) > > + + + + + > | | | | | > -> | | | | | > ____|_|_|_|_| > ____|_|_|_|/| > ____|_|_|_/|| > ____|_|_|/||| > ____|_|_/||||^ > ____|_|/|||||| > ____|_/|||||| > ____|/||||||| > ||||||||| > ||||||||| Actually, what you have drawn gives uniform phase but I understood your earlier descriptions. > Q1) Why do fringes form at all? Similar to Newton's rings but the separation is produced by the effect of motion on the path lengths. > Q2) If one beam is moved sideways by half a fringe, what happens to the > fringe > pattern? Nothing. I think you mean what happens if one image of the source is moved rather than the beam. In that case you start to change to parabolic fringes. > Q3) If the two beams reunite at not quite 90 degrees and their path > lengths are > slightly different, what happens to the fringe pattern. You would get vertical lines. This is why it is significant that the fringes are circular, it means both images of the source must lie on a line perpendicular to the screen and passing through the centre of the circular pattern but at different distances from the screen. > Q4) What other alternatives are there? There are many different interferometer designs out there but in the Sagnac experiment, the fact that the same mirrors participate in both paths means the beams are forced to be parallel giving circular fringes and when the equipment is rotated they remain circular but the radius changes as shown in the applet. http://www.physics.uq.edu.au/people/mcintyre/applets/michelson/michel.html Play around with it for a bit, it should help you follow some of what I am saying. George
From: George Dishman on 3 Dec 2005 05:08
"Henri Wilson" <HW@..> wrote in message news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... > On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > Frankly, I'm not quite sure how circular fringes are formed in either > interferometer. If the beams were both perfectly parallel and coherent, > there > should be no phase difference at any point across the whole viewing area. > I suppose if one considers that the two uniting wavefronts are spherical > rather > than flat, one gets a type 'Newton's rings' pattern...but why aren't the > wavefronts flat? Henri, I just realised you may be forgetting the effect of the eyepiece. Planes waves passing along the paths are focussed to a point by that. Otherwise there wouldn't be anything in the system to determine the centre of the circular pattern where you have an extended source and flat mirrors. George |