Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 23 Sep 2005 19:12 On Fri, 23 Sep 2005 13:47:47 +0200, "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote: >Henri Wilson wrote: >> On Tue, 20 Sep 2005 14:24:41 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)deletethishia.no> wrote: >> >> >>>You seem to be completely ignorant of how an >>>interference pattern is formed, and why fringes shifts. >> >> >> I'm not. > >Yes, you are. >Which you prove yet again in this posting. > >>>Please take the time to read the following properly, >>>I am using time to write it. >>> >>>To get an interference pattern with fringes, >>>the beams must be diverging and overlapping. >>> >>>Consider this simple figure: >>> >>> 1 2 >>> * * Two correlated (in phase) point sources >>> emitting monochromatic, coherent light. >>> (Laser and a beam splitter) >>> >>> >>> >>>--|--|--|---- screen >>> A B C >>> >>>The point B is equidistant to source 1 and 2. >>>We get a bright fringe through B. The fringe >>>will be a straight line. >>>The distance from the point A to point 2 is >>>half a wavelength longer than the distance >>>to point 1. We get a dark fringe through A. >>>This fringe will be a bit curved. >>>Likewise for point C, a dark fringe. >> >> >> Note that the angle to the observer's eyepiece is changing as you go from >> A->B->C. > >I typical Wilsonian nonsensical comment, and >a strong indication that you have no idea of >what you are talking about. > >You don't "go from A->B->C". >You don't move the screen, and you don't >move the eyepiece, so what the hell is your >comment supposed to mean? > >I would advice you to forget the eyepiece. >Think that the interference pattern is formed >on a screen. If the beam was perfectly parallel there would be NO inteference pattern on the screen. > >Using an eyepiece doesn't change the principle. >All it does is to project the light into your >eye so that the interference pattern is formed >on your retina. Have a think about interference from thin films. >>>Note that the reason why there are fringes >>>at all is that the beams are diverging, so >>>the distance from the source to the screen >>>is different on different parts of the screen. >>> >>>The angle of the beams when they unite >>>has obviously nothing whatsoever with >>>the matter to do. The only thing that matters >>>is the difference in the light path lengths >>>to the two sources measured in wavelengths. >> >> >> In tyhe sagnac, the two path lengths are different, whether or not each >> component's sped relative to the next IS considered. > >You are babbling. >I will take the typos as an indication that >you are writing without thinking. They are an indication that I am sick and tired of trying to educate you. >>>The only way to make the fringes move, >>>is to change the distance to one of the sources. >>>If we move source 1 a bit upwards, the fringes >>>will move to the left, and vice versa. >> >> >> If the two beams reunite at different angles, it is implied that they have >> traveled diferent lengths. > >Nonsense. >Read below. > >>>So when fringes move, the difference between >>>the two path lengths measured in wavelengths >>>changes. >>> >>>Now let this "interferometer" rotate. >>>Observed in the interferometer frame, >>>the light paths will be slightly curved, >>>so the angle with which the light hits >>>the screen is slightly altered. >>>But the fringes will not move, because >>>the slightly curved light paths from >>>point 1 to B and point 2 to B will still >>>be equally long. The angle at which the beams >>>hit the screen is utterly irrelevant. >> >> >> No. > >Yes. >Read below. > >>>And please don't say something like >>>"the beam will no longer hit point B, >>>because it is deflected." > >This IS what you are saying. >So read carefully: > >>>That is irrelevant. The beams are diverging >>>and overlapping, and what happens in point B >>>depend only on the lengths of the paths of >>>the light that hit point B, obviously. > >THINK about it. >Did you get it? > >>>The same applies for the four mirror set up. >>>It is stupid to say something like "the two >>>contra going beams will no longer combine >>>at the same point on the mirror." > >Henri, read carefully: > >>>Some light will always hit at "the midpoint" >>>of the combining mirror, and what happens in >>>that point is only determined by the phase >>>difference of the two light paths that >>>actually hit that point. If the fringes shifts, >>>it means that the phase difference changes, >>>which only can mean that the length difference >>>of the light paths have changed. >> >> >> Paul, George and I discused this at length. I have an animation of the way the >> beams move. I am content to believe that I won the argument. > >You have obviously not even understood how >an interefrence pattern is formed. Why do you think there are alternate light and dark rings, Paul. It is an ANGULAR thing. It is the angular difference that results in the path length difference. >I don't think you read my posting properly. >Please read it agin, carefully. too busy... >>>Sagnac falsifies the ballistic theory. >>> >>>No other conclusion is possible. >> >> >> Rubbish. > >It is the fact whether you understand it or not. > >"The ballistic theory predicts no length difference >of the light paths (measured in wavelengths)." >So far you agree. Absolute rubbish >"..and thus no fringe shifts when the Sagnac ring rotates." >Here is where you differ. > >But you cannot explain this away by claiming that >interferometers work in an entirely different >way than they actually do. > >Interferometers are used in a lot of different >applications, and they have been used for centuries. >It is very well know how they work. >Disputing this is idiocy. > >You are not even able to state what your alternative >explanation for how interferometers work, is. >You are only babbling some incoherent nonsense about >changing angles. It is all based on angles and the consequent path length diffferences at thsoe different angles. > >You are indeed living in Wonderland. > >To us in the real world, interferometers are no mystery. >We know how they work. ...and some of us know why they work. > >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".
From: Eric Gisse on 24 Sep 2005 00:38 Henri Wilson wrote: > On Fri, 23 Sep 2005 13:47:47 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >Henri Wilson wrote: > >> On Tue, 20 Sep 2005 14:24:41 +0200, "Paul B. Andersen" > >> <paul.b.andersen(a)deletethishia.no> wrote: > >> > >> > > >>>You seem to be completely ignorant of how an > >>>interference pattern is formed, and why fringes shifts. > >> > >> > >> I'm not. > > > >Yes, you are. > >Which you prove yet again in this posting. > > > >>>Please take the time to read the following properly, > >>>I am using time to write it. > >>> > >>>To get an interference pattern with fringes, > >>>the beams must be diverging and overlapping. > >>> > >>>Consider this simple figure: > >>> > >>> 1 2 > >>> * * Two correlated (in phase) point sources > >>> emitting monochromatic, coherent light. > >>> (Laser and a beam splitter) > >>> > >>> > >>> > >>>--|--|--|---- screen > >>> A B C > >>> > >>>The point B is equidistant to source 1 and 2. > >>>We get a bright fringe through B. The fringe > >>>will be a straight line. > >>>The distance from the point A to point 2 is > >>>half a wavelength longer than the distance > >>>to point 1. We get a dark fringe through A. > >>>This fringe will be a bit curved. > >>>Likewise for point C, a dark fringe. > >> > >> > >> Note that the angle to the observer's eyepiece is changing as you go from > >> A->B->C. > > > >I typical Wilsonian nonsensical comment, and > >a strong indication that you have no idea of > >what you are talking about. > > > >You don't "go from A->B->C". > >You don't move the screen, and you don't > >move the eyepiece, so what the hell is your > >comment supposed to mean? > > > >I would advice you to forget the eyepiece. > >Think that the interference pattern is formed > >on a screen. > > If the beam was perfectly parallel there would be NO inteference pattern on the > screen. Absolutely, without a doubt, wrong. The mathematics are trivial, and the observation of the effect is even more so. This only proves you have never even taken a basic physics course. > > > > >Using an eyepiece doesn't change the principle. > >All it does is to project the light into your > >eye so that the interference pattern is formed > >on your retina. > > Have a think about interference from thin films. An eyepiece is not a thin film. Have a think about what is required for thin film interference to occur. [snip] > > Why do you think there are alternate light and dark rings, Paul. > It is an ANGULAR thing. It is the angular difference that results in the path > length difference. Wow. Such arroange and ignorance, it is incredible. > > >I don't think you read my posting properly. > >Please read it agin, carefully. > > too busy... Again with "too busy". Odd how you always bring up how busy you are when you post dozens of times a day. [snip]
From: Henri Wilson on 24 Sep 2005 20:20 On 23 Sep 2005 21:38:56 -0700, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: >Henri Wilson wrote: >> On Fri, 23 Sep 2005 13:47:47 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)deletethishia.no> wrote: >> >> >I would advice you to forget the eyepiece. >> >Think that the interference pattern is formed >> >on a screen. >> >> If the beam was perfectly parallel there would be NO inteference pattern on the >> screen. > >Absolutely, without a doubt, wrong. > >The mathematics are trivial, and the observation of the effect is even >more so. > >This only proves you have never even taken a basic physics course. Geese, they don't teach the finer points in a basic physics course. I'll bet you wouldn't even know what a Bessel function was. >> >Using an eyepiece doesn't change the principle. >> >All it does is to project the light into your >> >eye so that the interference pattern is formed >> >on your retina. >> >> Have a think about interference from thin films. > >An eyepiece is not a thin film. No, but it views each part of the image from a slightly different angle. If the two beams of an interferometer were exactly parallel, what would determine where the fringes would appear? > >Have a think about what is required for thin film interference to >occur. You have no clue at all Geese. >> Why do you think there are alternate light and dark rings, Paul. >> It is an ANGULAR thing. It is the angular difference that results in the path >> length difference. > >Wow. Such arroange and ignorance, it is incredible. You have no clue at all Geese. > >> >> too busy... > >Again with "too busy". Odd how you always bring up how busy you are >when you post dozens of times a day. You have no clue at all Geese. > >[snip] 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: Eric Gisse on 24 Sep 2005 23:35 Henri Wilson wrote: > On 23 Sep 2005 21:38:56 -0700, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: > > >Henri Wilson wrote: > >> On Fri, 23 Sep 2005 13:47:47 +0200, "Paul B. Andersen" > >> <paul.b.andersen(a)deletethishia.no> wrote: > >> > > >> >I would advice you to forget the eyepiece. > >> >Think that the interference pattern is formed > >> >on a screen. > >> > >> If the beam was perfectly parallel there would be NO inteference pattern on the > >> screen. > > > >Absolutely, without a doubt, wrong. > > > >The mathematics are trivial, and the observation of the effect is even > >more so. > > > >This only proves you have never even taken a basic physics course. > > Geese, they don't teach the finer points in a basic physics course. Mabey, but they also don't teach points that are absolutely wrong. Such as the notion that two parallel beams of light do not produce interference patterns. > > I'll bet you wouldn't even know what a Bessel function was. What of them? Bessel function(s) are defined by an infinite series which is the solution to a particular ordinary differential equation with a parameter. If it served any purpose, I would reproduce the derivation of them for you. They are similar in feel to trig functions, in that they have an infinite number of zeros, and are bounded. Also J(0) can be 0 or 1 depending on the order of the function, and happens to oscillate somewhere inbetween -1 and 1. These come up, among other places, when solving the wave equation in three dimensions. So yes, I do know of them. What non-sequitur would you like me to address next? > > > >> >Using an eyepiece doesn't change the principle. > >> >All it does is to project the light into your > >> >eye so that the interference pattern is formed > >> >on your retina. > >> > >> Have a think about interference from thin films. > > > >An eyepiece is not a thin film. > > No, but it views each part of the image from a slightly different angle. Which has nothing to do with thin films. Thin film interference occurs when you have a film which has a size on the order of the wavelengths of light in question. Eyepieces do not apply because they are not thin, nor films, nor of high enough quality to be within a wavelength over the entire object. > > If the two beams of an interferometer were exactly parallel, what would > determine where the fringes would appear? If I had a scanner, I would scan the relevant page(s) from my introductory physics book. The entire chapter of light interference applies to this discussion. It is chapter 37, which starts on page 1178. The book is Physics for Scientists and Engineers, 6th ed, by Serway. In particular, you might want to pay attention to section 37.2 which describes Young's double-slit experiment [which you are arguing about, whether you know it or not]. Along with 37.6, which covers interference in thin films. > > > > >Have a think about what is required for thin film interference to > >occur. > > You have no clue at all Geese. Which is still far more clue than you have, considering you exhibit the symptoms of negative intelligence. > > > >> Why do you think there are alternate light and dark rings, Paul. > >> It is an ANGULAR thing. It is the angular difference that results in the path > >> length difference. > > > >Wow. Such arroange and ignorance, it is incredible. > > You have no clue at all Geese. > > > > >> > >> too busy... > > > >Again with "too busy". Odd how you always bring up how busy you are > >when you post dozens of times a day. > > You have no clue at all Geese. Why would I? You are always too busy to back up anything you say with facts! > > > > >[snip] > > > 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: jgreen on 25 Sep 2005 04:28
George Dishman wrote: > <jgreen(a)seol.net.au> wrote in message > news:1127473221.176343.176140(a)f14g2000cwb.googlegroups.com... > > > > George Dishman wrote: > >> <jgreen(a)seol.net.au> wrote in message > >> news:1127004753.829320.91010(a)z14g2000cwz.googlegroups.com... > >> > > >> > George Dishman wrote: > >> So far you haven't shown anything wrong with it. > >> If you want to do that, there have been two recent > >> papers which you could have cited that you haven't > >> even noticed, you are just going back to repeating > >> an error you made months ago when you misread a > >> post in the group. You seem to have forgotten you > >> got it wrong and only remember your glee at an > >> imagined problem. > > > > That would be the one where stars 8Gya old have been seen at 12.7 > > No such paper exists. > > > TERMINAL CANCER!!!!!!!!! > > Nope, it's the one that actually said the galaxies > were less than 2Ga old and 8Ga _away_ which you > misread as saying they were 8Ga _old_. Still that > suits your bias so I guess accuracy doesn't matter > to you as long as you can bolster your fantasy. > > George At what elapsed time after BB, did quasars form? What are their life expectancies? At what DISTANCE (age) have they been observed??? Jim G c'=c+v |