Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: donstockbauer on 21 Nov 2005 06:09 You are babbling, Henri. ********************** Perhaps he's the Star Child?
From: George Dishman on 21 Nov 2005 09:40 Henri Wilson wrote: > On Sun, 20 Nov 2005 10:23:12 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:nfhsn1p5qtslgismtsq4tr1akram78c641(a)4ax.com... > >> On Fri, 18 Nov 2005 15:31:03 -0000, "George Dishman" > > >> I should imagine that in a four mirror sagnac, fringe movement would be > >> quite > >> small. In a FoG with many turns, a much longer path length is provided and > >> hence much greater fringe movement. > > > >That's true, you can use more turns in the fibre > >coil, but it is still quite small. > > Are you sure? Yes, but it isn't too important as you'll see later, our disagreement lies elsewhere. Consider that Sagnac used an area of nearly a square metre and got a shift of 0.07 of a fringe at 120rpm. iFOGs tend to be in a box a few cm across and can just about sense the rotation of the Earth. That implies they are measuring about 10^-6 of a fringe at the low end up to maybe a few percent of fringe at their highest rate. > >> This is detected by focussing the fringe > >> pattern onto a narrow slit and counting the alternate dark/light > >> fluctuations. > > > >There is no slit or counter. A photodiode is used > >but the displacement is only a fraction of a fringe. > >The photodiode measures the brightness of the light > >which depends on the phase shift between the beams > >or to put it another way, which part of the pattern, > >a light or dark band, falls on the detector. > > that's what I meant. > It counts the light and dark pulses. It probably also senses 'fractions of > fringe.' My understanding is that they always work in the fraction of a fringe regime. The number of turns needed to work the way you suggest at low rates would be in the billions. However, if you understand 'counts the light and dark pulses' as just a way to extend the range beyond the fraction of a fringe region then it is functionally the same so as I said, not too important. You have to be careful not to get this out of place though, later you treat this counter as if it was integrating the signal, essentially you are introducing an extra integration into the electronics that isn't there If it worries you, the key might be for you to consider how you would determine the rate if you switch the equipment on while the vehicle is turning. You don't know how many counts away from zero you are at that time. > >> I'm not sure how fringe direction is monitored but it would be. > > > >That's actually the clever bit, the source is > >modulated so that the output is the product of a sine > >wave with the cosine produced by the interference. It > >means there is a sign change at the zero rotation point > >but gives the sensitivity of an interferometer. That's > >slightly different from the lab experiment version > >where fringes can be seen and counted as you suggest > >but the physical principles are the same, it's an > >engineering solution to the question of directional > >ambiguity. > > I don't quite see how the beam is modulated but I'll believe you. There are some web sites that give a rough outline but I've only seen one with any details, I'll post it if I can find it again. Basically they use a piezo crystal clamped onto the fibre. > >Given that the displacement is a fraction of a fringe > >and generally has a sine curve, the current in the > >photodiode is roughly proportional to the fringe > >displacement, or technically to the phase difference > >between the two beams. > > > >Phase difference is 2 * pi times time difference > >over the period or 2 * pi * frequency * delta_t. > > > >The shift is actually much smaller but: > > > >> Now, during an acceleration period, 50 such pulses might be counted. > > > >The displacement is proportional to the acceleration > >so "As the acceleration smoothly increases, 50 such > >pulses might be counted." > > > >> When the > >> acceleration ceases, ... > > > >"As the acceleration smoothly decreases, 50 such > >pulses would be counted but in the opposite > >direction." so that, once the speed returned to > >zero, the counter would also read zero. > > It would read what it read before the acceleration, not zero. Yes but see next. > >> ... the visible fringe might not be exactly in the centre of > >> the slit > >> and I assume there is a way of determining that offset...because it > >> is essential for accuracy..... > > > >The offset would have been present before the > >acceleration and could be calibrated out in > >the factory. Only long term drift or temperature > >sensitivity causes a problem. > > By 'offset' I meant 'fringe displacement'. By 'offset' I meant 'the residual measured fringe displacement when the angular speed is actually zero'. It will always exist due to voltage offsets in DC amplifiers and leakage on AC signals so will need to be cancelled out by some calibration procedure. I think if you look back at the last two paragraphs, we are really saying the same thing. > >> since the offset is integrated electronically > >> with time to obtain total rotation angle. > > > >No, the output from what you say above if integrated > >would give speed since it is proportional to > >acceleration. > > Wait I think we are talking about different things here George. Sort of. > The integration of fringe displacement with time gives total angle moved during > that time. That's right, but also the integration of acceleration would give the speed. Where we disagree, perhaps, is whether the fringe displacement is a function of the speed, the acceleration or a bit of both. Again, this will become clearer later. > >> I think you and Andersen are omiting a factor of (2-Root2) when you claim > >> that > >> the displacement should return to zero during constant rotation, according > >> to > >> the BaTh...In fact, the travel time around each path is not the same. > > > >A factor of root two applied to zero is still zero > >but I am prepared to consider there might be a root > >two scale factor difference, I haven't looked at it > >in detail. There is a fundamental difference between > >an output proportional to speed as predicted by SR > >and an output proportional to acceleration as > >predicted by Ritz which makes a factor of 0.7 of > >lesser significance at the moment. > > The output, as in 'fringe displacement' is still proportional to rotation speed > according to Ritz. That's where we disagree. See below. > >>>Correct, by a constant amount while the acceleration is > >>>present, and since it is proportional to the acceleration, > >>>it goes back to zero if the speed subsequently becomes > >>>constant again. > >> > >> Haha. > >> So you apply an acceleration that slowly approaches zero. During that > >> time, the > >> fringes move in one direction only. > > > >Yes, they slowly approach zero because their > >displacement is proportional to the acceleration. > > No george. Their CHANGE of displacement occurs DURING acceleration. > They slowly approach their new position. Again see below. > >> Do you really think they suddenly flip right back to the starting point > >> when > >> the acceleration ceases? > >> > >> That's nonsense George. > > > >It is indeed. > > Good. > > > > >>>> My diagram shows why. The path length change only during acceleration, > >>>> not > >>>> during constant rotation. > >>> > >>>Careful with your wording Herni, your diagram shows the > >>>path length ARE CHANGED during constant acceleration from > >>>their values during constant velocity. > >> > >> The path lengths remain constant during constant rotation. > > > >And the speeds remain constant at c+v and c-v so the times > >are equal. > > what about the (2-root2) factor? Consider it within the calculation of v. You need to look at the duck, the car and the goose to resolve the quantitative aspect but that's probably of lesser importance at the moment. > >> During a period of acceleration, I think the path length is CONTINUALLY > >> increasing. > > > >That's not what your diagram shows. For CONSTANT > >acceleration, the path lengths are constant and > >changed from the non-accelerating length by an > >amount you show as 1/2 a t^2. > > > >> I too have been working on that point. It's not all that simple. > > > >Trust your diagram then. > > You are probably right. > ...but you seem to believe the path lengths revert to their old values after > the acceleration ceases. > They don't. _That's_ where we differ ! > Each rotation speed has a different path length. You are ignotring the 'vt' > term. Not ignoring it, the path length changes as you say but the your diagram is drawn in the lab frame so the speed is not c, it is also modified. That was the point I have been making for months, the speed and path length changes to the two paths are equal but opposite (+vt versus -vt, c+kv versus c-kv) so that factor cancels. You are right about the acceleration though and I though that was what you had realised a week or so ago when you posted: >"Henri Wilson" <HW@..> wrote in message >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... >> >> George, George, George. >> >> I have finally woken up to your (and MY) complete misinterpretation >> of the problem. >> >> We have both been arguing about whether or not the fringes will >> move during constant angular rotation...and of course they don't. > >>>> The fringe 'displacement' at any instant is the integrated effect of all > >>>> previous ACCELERATIONS. > >>> > >>>There is no physical mechanism involved that could integrate > >>>the difference in arrival times of wavecrests. The actual > >>>path times don't directly produce an output. > >> > >> During an acceleration, the two beams continuosly 'beat'. Even Androcles > >> get's > >> it right occasionally. > > > >During a _change_ of acceleration, they beat. When > >the acceleration becomes constant, the stop beating > >at take a stable value of phase shift at the value > >they had at that time. > > That's right...meanwhile the number of fringes moved (including the partial bit > at the end) has been counted. Careful, as I said above I don't believe there is a counter, just the fractional part detector, but let's look at what would happen if there were a counter. As the acceleration build up to the constant value, there is a beat which is counted, the fractional part being added. (The counter would increment each time the fractional part rolled over.) While the acceleration is constant, the displacement is constant so the counter output is constant. While the acceleration decreases to zero, the fringes would pass in the other direction so the couter would need to decrease until it should be back at zero when the rotational reaches zero. The output of that counter would therefore be a digital indication of the acceleration, not the speed. One point Henri, in this part you seem to have lost the "vt" term which you mention above. I think you need to consider just which of the terms is responsible for the output here, it's quite fundamental. > >> The beats are counted. > > > >That would be an electronic integration which would > >give speed. Actually I was wrong there, the first counter gives acceleration, a separate integrator is needed to get speed and a second to get angle turned. > It gives change in speed. Both integrators need to have an initial value placed in them. > A second continuous integration of instantaneous displacement with time gives > the rotation angle from zero. Exactly. > >You could imagine that this is how commercial units > >work (it isn't) but that wouldn't apply to the lab > >experiment. Remember in the original experiment of > >Sagnac, he saw a shift of 7% of a fringe while the > >table was turning at 120rpm. What you are describing > >would be that he counted 0.07 fringes during the > >acceleration phase and the displacement returned to > >zero once constant speed was achieved. > > I dont think that would be accurate enough for any practical purpose. > I'm sure fringes move a lot more in multi turn FoGs. Well see the numbers above. > >> The count IS really an integration of the path length > >> change (or, more correctly, the number of wavelengths in each path) > > > >Yes but counting would have to be done as part of the > >electronics, it is not inherent in the physical process. > > The path length change effectively integrates acceleration. No need for > electronics. That's not what you described above where the fringe counter is supposedly doing the integration. > >> The count itself (fringe diplacement) is also electronically integrated to > >> give total rotation angle. > > > >That would be a second integration. > > yes. I think we are thinking along very similar lines, the difference between us is really in the detail, unless you want to go back to sayig the acceleraton term is a transient and it is the "vt" part on your diagram that is responsible for the output. George
From: Paul B. Andersen on 21 Nov 2005 17:48 Henri Wilson wrote: > On Thu, 17 Nov 2005 16:54:34 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Henri Wilson wrote: >> >>>On Wed, 16 Nov 2005 23:51:35 +0100, "Paul B. Andersen" >>><paul.b.andersen(a)hiadeletethis.no> wrote: > > >>>>Look Henri. >>>>Pick a distant star, say 500 LY away. >>>>Point your telescope at it, so that the image is at the centre. >>>>Measure the absolute angle of your telescope. >>>>Repeat 6 month later. >>>>The telescope will now point in a direction 22 arcsecs >>>>different from the first time. >>>> >>>>The parallax is negligible. The light path is the same, >>>>nameley a straight line from the star to the Earth. >>>>So why are the angle of the light path different? >>>>It is caused by the different velocity of the Earth >>>>at the two occations. We are observing the star from >>>>two different frames of reference. >>>>Both frames are moving relative to the star. >>> >>> >>>So what? Ligth leaves the star spherically. >> >>Still drunk? >>Didn't you get it? >>There is but one light path - the path from the Star to the Earth. > > > Are you under the impression that the star is emitting all its light in one > particular direction, as with a narrow laser beam? Still drunk? Or are you too stupid to get the simple point? The light that does not hit the CCD in our telescope is of no interest whatsoever. It might as well have been a laser beam, it would make no difference. (The width of the beam would have to be at as wide as the parallax angle, though. But as this is much smaller than the aberration angle, we can forget it.) Ignore the parallax. We can in our thought example imagine that the star is in the equatorial plane, and that we are observing it when the Sun, Earth and star are all in line. (That means that the star will be behind the Sun at one occasion, but it doesn't matter in principle) Then we have the picture drawn in solar frame: * star | | the one and only | light path | | O-> v observer | S | v<-O At one occasion, the observer is moving to the right, an will have to point his telescope like this: / / light path down the tube / ------ Six month later, the observer is moving to the left, and will have to point his telescope like this: \ \ light path down the tube \ ------ At both occasions, the angle is v/c radians from vertical. v/c = 10^-4 radians = 11 arcsecs The difference is 22 arcsecs. > >>(We can neglect the small parallax angle which is only 0.0006 of >> the aberration angle) >>It is obviously utterly irrelevant that the star emits light >>in all other directions that don't hit the Earth, >>so why the hell are you stating this stupidity? >> >>The only light path of interest is the one that hits the Earth! >>The _direction_ of that light path is down the middle of >>our telescope. > > > You really are funny today. > > >>The direction of that single light path changes throughout the year >>because the velocity of the frame of reference (Earth) changes >>throughout the year. > > > Very good Paul. You are improving. > > >>>This is in no way related to our discussion. You are diverting attention from >>>the fact that SR is proved to be nonsense. >> >>It is related to your incredible stupid statements: >>"Wavefronts really exist only in the source frame." >>and: >>"Whatever is moving diagonally isn't light. It is >>infinitesimal points." > > > That is right. Of course I was refering to the plotting, in my frame, of the > paths of individual 'points' inside a vertical laser beam as I move > horizontally past it. You are evading the point, Henri. There is no important difference from your vertical beam. The observer is moving perpendicular to the light beam from the star that hits the Earth. That light path from the star is "vertical" in our "stationary" solar reference system, which is the same as the "source frame". But the light path that is going down the observer's telescope is at an angle v/c radians from vertical. So what is it that is moving at an angle down the tube and hitting the CCD? Is it light, or is it just infinitesimal points? How can the CCD detect infinitesimal points? > George Dishman is too dumb to understand that but I thought you might have a > little more sense. > > >>To be consistent, you have to claim: >>"WHATEVER IS COMING FROM THAT STAR, MOVING WITH CHANGING >> DIRECTION, IS NOT LIGHT, IT IS ONLY INFINITESIMAL POINTS." > > > You are quoting me completely out of context and you know it. > tat has nothing whatsoever to do with the topic or anything I have said. No, I am not. You claimed that the light is moving vertically in the "source frame", and "whatever is moving diagonally in the observer frame is not light." The light from the star IS moving vertically in the source frame. But our observer have to point is telescope "diagonally" to make the "whatever that is moving" go down his telescope tube. So what is the "whatever" that is moving diagonally down the tube? You claim it is not light. So what is it then? Whatever it is, CCDs detects it as if it were light. >>Now Henri, what is it that hits the CCD in our telescope? >>Is it light? >>It cannot be, can it? >>Because whatever moves along paths with different directions >>in different frames cannot be light, can it? > > > The star emits a sphere of light Paul. The wavefronts are spherical. Didn't you > know that. > When my telescope moves sideways, a different radius vector of the sphere goes > down the middle of my telescope. What could be more simple? I cannot see why > you should have any trouble understanding that. Please, Henri. You are not really THIS stupid, are you? :-) (I am beginning to believe you are. I didn't think it possible!) Don't you understand that if you had two telescopes going in opposite direction, looking at the same star, they would have to point in opposite directions as they pass each other? * | 100 LY | / \ / \ O->v<-O It is the very same one and only light path that has different angles in the two observers' frames. A little movie for you. The "*" is a photon, or a bit of LIGHT. Two telescope tubes going through each other in opposite directions. enter X * X tubes / \ / \ / X \ / / \ \ / / \ \ -ccd- -ccd- \ / \ / X * X going / \ / \ down / X \ / / \ \ -ccd- -ccd- \ X / and \ / \ / down X * X / \ / \ / X \ -ccd-ccd- \ \ / / \ X / \ / \ / X * X / \ / \ -ccccd- \ \ / / \ \ / / \ X / \ / \ / X * X -ccd- \ \ / / \ \ / / hit \ \ / / ccds \ X / \ / \ / Xc*dX Our single photon - or piece of LIGHT - goes down both tubes. This single photon has obviously but one path. This single light path is vertical in the screen frame, AND the very same light path is "diagonal" in the two telescope frames. The photon - or piece of light- is moving vertically in the screen frame, AND the very same photon - or piece of light- is moving diagonally in the telescope frames. It is obviously mindless babble to claim that the very same physical object both is and isn't light. Paul
From: Henri Wilson on 21 Nov 2005 18:12 On 21 Nov 2005 06:40:55 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >Henri Wilson wrote: >> On Sun, 20 Nov 2005 10:23:12 -0000, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >> >That's true, you can use more turns in the fibre >> >coil, but it is still quite small. >> >> Are you sure? > >Yes, but it isn't too important as you'll see later, our >disagreement lies elsewhere. Consider that Sagnac >used an area of nearly a square metre and got a shift >of 0.07 of a fringe at 120rpm. iFOGs tend to be in a >box a few cm across and can just about sense the >rotation of the Earth. That implies they are measuring >about 10^-6 of a fringe at the low end up to maybe a >few percent of fringe at their highest rate. I can't find a decent description of a FoG anywhere. I was under the impression that path lengths were pretty high, maybe 1000 metres or so.. > >> >> This is detected by focussing the fringe >> >> pattern onto a narrow slit and counting the alternate dark/light >> >> fluctuations. >> > >> >There is no slit or counter. A photodiode is used >> >but the displacement is only a fraction of a fringe. >> >The photodiode measures the brightness of the light >> >which depends on the phase shift between the beams >> >or to put it another way, which part of the pattern, >> >a light or dark band, falls on the detector. >> >> that's what I meant. >> It counts the light and dark pulses. It probably also senses 'fractions of >> fringe.' > >My understanding is that they always work in the >fraction of a fringe regime. The number of turns >needed to work the way you suggest at low rates >would be in the billions. I cannot see how any kind of acceptable accuracy would be achieved if only fractions of fringes were measured. I don't know how the source would be kept sufficiently stable for one thing. Also the detector would be quite temperature sensitive. > >However, if you understand 'counts the light and >dark pulses' as just a way to extend the range >beyond the fraction of a fringe region then it is >functionally the same so as I said, not too >important. You have to be careful not to get this >out of place though, later you treat this counter as >if it was integrating the signal, essentially you are >introducing an extra integration into the electronics >that isn't there > >If it worries you, the key might be for you to consider >how you would determine the rate if you switch the >equipment on while the vehicle is turning. You don't >know how many counts away from zero you are at >that time. Point taken. It would have to be on all the time...or reset from some arbitrary zero. >> It would read what it read before the acceleration, not zero. > >Yes but see next. > >> >> ... the visible fringe might not be exactly in the centre of >> >> the slit >> >> and I assume there is a way of determining that offset...because it >> >> is essential for accuracy..... >> Wait I think we are talking about different things here George. > >Sort of. > >> The integration of fringe displacement with time gives total angle moved during >> that time. > >That's right, but also the integration of acceleration would >give the speed. Where we disagree, perhaps, is whether >the fringe displacement is a function of the speed, the >acceleration or a bit of both. Again, this will become >clearer later. I say the fringe displacement (noun) is a function of speed. Changes in displacement occur DURING acceleration. The rate of change is a function of acceleration. >> >> The output, as in 'fringe displacement' is still proportional to rotation speed >> according to Ritz. > >That's where we disagree. See below. >> >> No george. Their CHANGE of displacement occurs DURING acceleration. >> They slowly approach their new position. > >Again see below. >> >> I too have been working on that point. It's not all that simple. >> > >> >Trust your diagram then. >> >> You are probably right. >> ...but you seem to believe the path lengths revert to their old values after >> the acceleration ceases. >> They don't. > >_That's_ where we differ ! > >> Each rotation speed has a different path length. You are ignotring the 'vt' >> term. > >Not ignoring it, the path length changes as you say but the >your diagram is drawn in the lab frame so the speed is not >c, it is also modified. That was the point I have been making >for months, the speed and path length changes to the two >paths are equal but opposite (+vt versus -vt, c+kv versus c-kv) >so that factor cancels. I think you are trying to say that even though the path lengths of the two beam change during acceleration and remain changed by a constant amount during constant rotation, the travel time of light in each beam is always the same. I say it is slightly different but constant.... and so there is a no fringe movement during constant rotation. The number of wavelengths in each path is different (you claim only fractionally, no matter) >You are right about the acceleration >though and I though that was what you had realised a week >or so ago when you posted: > > >>"Henri Wilson" <HW@..> wrote in message >>news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... >>> >>> George, George, George. >>> >>> I have finally woken up to your (and MY) complete misinterpretation >>> of the problem. >>> >>> We have both been arguing about whether or not the fringes will >>> move during constant angular rotation...and of course they don't. > > >> >>>> The fringe 'displacement' at any instant is the integrated effect of all >> >>>> previous ACCELERATIONS. >> >>> >> >>>There is no physical mechanism involved that could integrate >> >>>the difference in arrival times of wavecrests. The actual >> >>>path times don't directly produce an output. >> >> >> >> During an acceleration, the two beams continuosly 'beat'. Even Androcles >> >> get's >> >> it right occasionally. >> > >> >During a _change_ of acceleration, they beat. When >> >the acceleration becomes constant, the stop beating >> >at take a stable value of phase shift at the value >> >they had at that time. >> >> That's right...meanwhile the number of fringes moved (including the partial bit >> at the end) has been counted. > >Careful, as I said above I don't believe there is a counter, >just the fractional part detector, but let's look at what >would happen if there were a counter. > >As the acceleration build up to the constant value, there >is a beat which is counted, the fractional part being added. >(The counter would increment each time the fractional part >rolled over.) While the acceleration is constant, the >displacement is constant so the counter output is constant. No George. You are confusing constant acceleration with constant speed. >While the acceleration decreases to zero, the fringes would >pass in the other direction so the couter would need to >decrease until it should be back at zero when the rotational >reaches zero. No that is not right. If the acceleration is +ve, the fringes move one way. It matters not whether the magnitude of that acceleration is increasing or decreasing. You are claiming the fringe ''''movement'''' is a function of da/dt. I would like to see your proof. >The output of that counter would therefore be a digital >indication of the acceleration, not the speed. > >One point Henri, in this part you seem to have lost the "vt" >term which you mention above. I think you need to consider >just which of the terms is responsible for the output here, >it's quite fundamental. The at^2/2 is responsible for fringe movement. The vt is responsible for fringe displacement during constant rotation. > >> >> The beats are counted. >> > >> >That would be an electronic integration which would >> >give speed. > >Actually I was wrong there, the first counter gives acceleration, >a separate integrator is needed to get speed and a second to >get angle turned. > >> It gives change in speed. > >Both integrators need to have an initial value placed in them. yes > >> A second continuous integration of instantaneous displacement with time gives >> the rotation angle from zero. > >Exactly. > >> >You could imagine that this is how commercial units >> >work (it isn't) but that wouldn't apply to the lab >> >experiment. Remember in the original experiment of >> >Sagnac, he saw a shift of 7% of a fringe while the >> >table was turning at 120rpm. What you are describing >> >would be that he counted 0.07 fringes during the >> >acceleration phase and the displacement returned to >> >zero once constant speed was achieved. >> >> I dont think that would be accurate enough for any practical purpose. >> I'm sure fringes move a lot more in multi turn FoGs. > >Well see the numbers above. Like I said, I can't find a decent description of FoGs and their design features. .....Nothing much on google. > >> >> The count IS really an integration of the path length >> >> change (or, more correctly, the number of wavelengths in each path) >> > >> >Yes but counting would have to be done as part of the >> >electronics, it is not inherent in the physical process. >> >> The path length change effectively integrates acceleration. No need for >> electronics. > >That's not what you described above where the fringe >counter is supposedly doing the integration. It registers the answer. > >> >> The count itself (fringe diplacement) is also electronically integrated to >> >> give total rotation angle. >> > >> >That would be a second integration. >> >> yes. > >I think we are thinking along very similar lines, the difference >between us is really in the detail, unless you want to go back >to sayig the acceleraton term is a transient and it is the "vt" >part on your diagram that is responsible for the output. See my comment above about da/dt Let's clear that up. > >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: Henri Wilson on 21 Nov 2005 18:29
On Mon, 21 Nov 2005 11:51:10 +0100, "Paul B. Andersen" <paul.b.andersen(a)hiadeletethis.no> wrote: >Henri Wilson wrote: >> On Fri, 18 Nov 2005 14:08:21 +0100, "Paul B. Andersen" >> <paul.b.andersen(a)hiadeletethis.no> wrote: >> >> >>>>Wouldn't you like that to be true eh? >>>> >>>>At what tempertatures do they run? >>> >>>Depend on the cooling. >>>Say 50 - 150C. >>> >>> >>>>What determines the direction of emission of a photon from an atom Paul? >>>>Come on, you are the expert..... >>> >>>I am a little reluctant to use much time to explain >>>to you what you will ignore anyway. >> >> >> I take that to mean you don't know. >> ..which isn't surprising because I doubt if anyone has the faintest idea. > >I didn't say I couldn't or wouldn't explain it. >I said I was reluctant do to so, because you >would probably not care to read it anyway. >Which you have demonstrated below. > >>>But in a laSEr, we are talking about Stimulated Emission. >>>That means that a photon passing close by an atom may stimulate >>>it to emit a new photon with the same direction and with the same >>>phase. >> >> >> Well now, that IS interesting. >> It appears to support my W-aether theory. >> >> >>>>Why should gas lasers work at all? >>> >>>Why indeed. :-) >>>I am not going to give a thorough description, you can look it up >>>yourself if you really want to know. >>>But you never really want to know, you will rather invent it yourself. >>> >>>But I will mention a few points of special interest to this discussion. > >But Henri didn't read it, and made no attempt to understand it. > >>>Let us consider a HeNe laser. Let us assume the gas temp. is 350k. >>>Let us assume the length of the tube is ca. 0.75 m. >>>The rms speed of the He atoms will be: >>>v^2 = 3kT/m where m is 4 proton masses. >>>v = 1.47 km/s. (rms) >>>v/c = 5*10^-6. >>>This means that the frequency emitted by the atoms >>>will be Doppler shifted, so the frequency will be >>>distributed like a Gauss function with relative width ca. 5*10^-6. >>>The central frequency fo = ca. 0.5*10^15 Hz >>> >>>However, the laser tube is a resonator, and only the frequencies >>>given by f = m*c/2L can exist in the resonator. >>>The distance between the possible frequencies is: >>>delta_f = c/2L = 2*10^8 Hz. >>>delta_f/fo = 4*10^-7 >>> >>>Note that a laser do not emit strictly monochromatic >>>light, but a number of close spectral lines. >>>We can see that in the order of 10 spectral lines can exist >>>within the frequency distribution above. >>> >>>We KNOW this is happening in a laser. >>>Each spectral line is emitted by atoms with >>>a specific longitudinal velocity component. >>> >>>According to the BaT, the light from these spectral >>>lines should travel at different speeds. >>>They don't. >>>BaT falsified. >> >> >> Yes, very funny Paul. >> >> I think it is a little more complicated than that. > >So what is too simple to you, Henri? :-) How is your 'm' in 'v^2 = 3kT/m' related to your 'm' in 'f = m*c/2L' >> I think the photon fields combone somehow to create the resonance. > >Of course they do. >They combine to a number of standing waves in the resonator. >That WAS what I explained above, wasn't it? No . How are your m's connected? >> I don't think molecular speeds would make much difference. > >You are babbling, Henri. >The distribution of atomic speeds is the very reason for >why there are several close spectral lines in a laser beam. >And the number of spectral lines is exactly as predicted >by the speed distribution. >Read the explanation above again, please. At best, your explanation was very unclear. At worst plain bulldust. >The light in one of these of spectral lines can only come >from He atoms with exactly the right speed to Doppler >shift the light by exactly the correct amount. >So we know what the speed of the atoms emitting the light >in each spectral lines is. >To sum it up: >1. We know that the speed of the atoms emitting the light > in the different spectral lines are different. >2. We know that the speed of the light in the different > spectral lines is the same. > >The BaT is falsified. >Again. ....and laser beams are infinitesimally bright according to Andersen's theory. > > Remember it was YOU > > who claimed that electric fields act instantaneously. > >I claimed that a force acts on a charged particle >at the same instant it enters a static electric field. >You claimed this could be used to instant communication! :-) > >The fact that you mention this utterly irrelevant matter >show your inability to give any sensible respons to the issue >at hand. Sort yourself out Paul. > > >Paul 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". |