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| Laserlight |
Posted: Feb 12 2007, 04:38 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
Hi C2, I liked your analogy!
"Clues" lead you toward the elephant, and elephants leave "huge clues", but just stepping in the obvious "clues" doesn't mean that you have actually found the elephant. If every "clue" looks the same, smells the same, and leaves the same residue on your shoes that still doesn't mean it is the true "essence" of the elephant. Claiming the elephant is nothing more than the "clues" it leaves is like opening a wrapped birthday gift to find nothing inside the box. It looks great on the outside with lots of pretty wrappings and ornate ribbons and bows, but there is nothing of substance on the inside. Please take off your shoes outside...... LL |
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| Laserlight |
Posted: Feb 12 2007, 04:56 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
C2,
Yes it does, that is why the pattern changes when the slit width or shape changes. You have changed the "dynamics" of the geometry of the slits with the geometry of the impinging light. If the length of a waveguide changes the output will remain the same but will take longer to pass thru the length of the waveguide, because the shape of the pulse energy travelling thru it will assume some average distribution (standing waves) over the distance of the waveguide, but if the width of the waveguide changes the output will have a different shape relative to the output from the original waveguide. The external x and y dimensions of the waveguide are what control the shape of the projected pattern. The z dimension controls the delay of the signal moving thru the waveguide. The time travelling thru the waveguide will take longer than the time for the wave travelling in open space over the same distance. Regards, LL This post has been edited by Laserlight on Feb 12 2007, 04:58 PM |
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| Montec |
Posted: Feb 12 2007, 05:06 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 807 Joined: 9-November 05 Positive Feedback: 60.71% Feedback Score: 19 |
Hello all
If the Michelson interferometer makes you scratch your head then this will cause you to pull out your hair. http://www.phy.duke.edu/ugrad/thesis/crawf...wfordThesis.pdf Have a look. -------------------- Competition is the essence of evolution.
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| TRoc |
Posted: Feb 12 2007, 06:52 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 913 Joined: 5-October 04 Positive Feedback: 80.77% Feedback Score: 26 |
Hi all, A side note: thanks to C2's analogy, we now have this "smart ad" appearing at the bottom of this page:
What a great world we live in! C2, your reluctance to change is amazing, and your logic incomprehensible. I mentioned "wavefronts" reforming a-la Huygens' method in the post prior to last, with the caveat that the reforming, expanding wavefront will hit the wall continuously. That means WITH EQUAL PATH LENGTHS from "new" (ad hoc) point sources. The FACT is, you CAN NOT use BOTH of these analogies: it is one or the other. MY main goal is to get everyone into their respective corners of "belief". It's great that you have dropped the "particle" approach to the "photon". If you are now going to switch to the Huygens' method (from the one you have consistently been proposing), then guess what? ... you MUST also accept that "photons" ARE able to interact in free space, as the method requires. You have consistently denied this possibility. In fairness to the rest of us, you should choose a position, and stick with it. There is nothing wrong with changing your mind, but, if it contradicts your prior position, you should inform everyone so we know what arguments to use. The good news is that regardless of which of these methods you prefer, we will still need a physical explanation for the results. "Predicting" that my car will start when I turn the key is lifetimes away from explaining WHY. To do that would require a cohesive, full understanding of several different systems, and "phenomena". Even if we "keep it simple", and just admit Nitrogen and Oxygen into the real medium of the DSE, the "randomness" and degrees of freedom of these elements completely deny any "ordered structures" from existing. "Ordered structures" would include both "equidistant points across the slit width", and "new point sources at the symmetrical junction of spherical wave fronts". LL was nice enough to point out the math of what the size of the bands should be according to theory, and didn't even divide by 2 (at 1/2 wavelength "destructive" interference). Simple point being: the diffraction pattern looks nothing like this: |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||| with 350nm separation (that would render it "invisible" to the naked eye), using his 700nm light source analogy. That is why I suggested the "molecule as compass" analogy. The diffraction pattern is many orders of magnitude larger than sub-wavelengths required by your model. To the eye, this would just be a solid band of "gray" light. (not bright, not dark) The real pattern has "speckle interference" between the light and dark bands, and is NOT clearly separated. There is an 5% accepted "noise" level in the pattern, that precludes sharp contrast lines that your method would produce. Perhaps most importantly, if your model was "perfect" as you claim, Quantum Mechanics would not exist as we know it. It is YOU who needs to defend the position that "nothing else" is required for an explanation, NOT us. Historically, Science had to move PAST these first several attempts at explaining the results, BECAUSE THEY WERE INADEQUATE. You act like we are still in the early 20th century, and QM doesn't exist. The PROOF that "photons" are hitting the screen at the dark band locations is SIMPLE, and has been done. You just remove the screen, and insert a NEW slit, in the EXACT location where the dark band just was, and guess what? YOUR model predicts that NOTHING would pass through the new slit, yet, A NEW INTERFERENCE PATTERN EMERGES, from the light diffracting through the slit. I recall that you stated earlier that "energy" was "shifted" from the dark bands to the light bands, at the screen. This contradicts your much professed model as well. Please give up, you have no hope of overturning the overwhelming evidence that requires a more complex model than is used in introductory, high school Physics. regards, T.Roc -------------------- I know Nothing. I looked all over to find it, but found it Nowhere. The funny thing is, it was right between 2 things, that I knew Everything about. It felt like forever, but really, it was no Time at all. |
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| Laserlight |
Posted: Feb 12 2007, 07:42 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
TRoc, As much as it pains me to disagree on a couple of points, I feel that I must....
Without perfect spacial and temporal coherence, light cannot interact until it "interferes" with the EM fields of matter that function as a "mixing" catalyst to combine and unite the temporal and spacial factors at a common reference point.
So what do you suppose happens to the DSE carried out in a vaccum? I am proposing that the results will be the same. Do you disagree? Regards, LL |
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| Confused2 |
Posted: Feb 12 2007, 07:42 PM
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Toothpaste salesman ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 4755 Joined: 8-November 05 Positive Feedback: 63.3% Feedback Score: -31 |
Just a recap for anyone who has recently joined this rather long thread ..
This is the experiment we have looked at in some detail.. 1/ http://www.teachspin.com/instruments/two_s...periments.shtml In particular .. this is the graph of 'that which is to be explained... 2/ http://www.teachspin.com/instruments/two_s..._combiplot2.gif The amazing ripple tank is good for getting some grip on the problem 3/ http://www.echalk.co.uk/Science/Physics/ri...Tank/ripple.htm And here we have two references which I have given many times.. 4/ http://www.hmi.de/bereiche/info/dualismus/exp.java_en.html This is the one known as 'The DSE equation' 5/ http://schools.matter.org.uk/Content/Inter...ce/formula.html Please feel free to join in! Best wishes, -C2. |
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| Laserlight |
Posted: Feb 12 2007, 08:15 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
C2,
Look closely at your wave tank. Where is the signal mixing taking place? As I look closely at the striated wave fronts emanating from area of the 2 slits, I can follow them back to the area immediately between the 2 slits which is the mixing point of the spacially separated wave front signals. Timing wise, the wavefronts are virtually the same, but they are spacially separated by the gap between the slits. The signals are mixing/combining and interfering between the slits. From that point the mixed and interfered waves just radiate and expand outwards toward the detection screen, which is just a projection angle from the focal point where the mixing takes place. The light and dark bands are merely projections of the interference mixing and focal actions that take place at the mixing point. Other comments, opinions, observations, disagreements, welcomed. LL This post has been edited by Laserlight on Feb 12 2007, 08:17 PM |
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| Confused2 |
Posted: Feb 12 2007, 08:22 PM
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Toothpaste salesman ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 4755 Joined: 8-November 05 Positive Feedback: 63.3% Feedback Score: -31 |
Hi TRoc,
I have posted the DSE equation many times and yet you don't seem to have looked at it even once. The classic DSE equation was the breakthrough that enabled Thomas Young to find the wavelength of light Since we now know the wavelength of light We have to recast Wavelength = (Distance between peaks) x (slit separation) / (Distance to sceen) To the form (Distance between peaks) = (Wavelength) x (Distance to screen) / (Slit separation) If the wavelength is (say) 700nm and the distance to the screen is (say) 700mm and the slit separation is (say) 0.5mm then (Distance between peaks) = 700x10^-9 x 700x10^-3 / ( 0.5x10^-3) = 980,000 x 10^-6 = 0.9mm .. Since I have chosen (from memory) roughly the dimensions of the Teachspin setup we get http://www.teachspin.com/instruments/two_s..._combiplot2.gif and we see the bright bits are abot 0.9mm apart. From the DSE equation. We predict, we look and we find that it is good.. It doesn't matter if I've got the dimensions slightly wrong . the equation works every time The path length difference is (correctly) predicting how far apart the fringes .. please look at the drawing ( http://schools.matter.org.uk/Content/Inter...ce/formula.html ).. just once. It's virtually the same as the one for diffraction... no new concepts. Best wishes, -C2. |
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| Laserlight |
Posted: Feb 12 2007, 08:43 PM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
C2,
So how does your model account for the single photon interference result? Can you explain that, since photons are not mixing at the screen but have a pattern of distribution. Where does your calculation of signal mixing overlap take place? LL This post has been edited by Laserlight on Feb 12 2007, 08:47 PM |
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| Confused2 |
Posted: Feb 12 2007, 09:58 PM
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Toothpaste salesman ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 4755 Joined: 8-November 05 Positive Feedback: 63.3% Feedback Score: -31 |
Hi TRoc, Laserlight et al,
It has been a major mistake (of mine) to imagine we could usefully discuss the single photon result before we have reached any agreed understanding of the classical wave phenomena which produce interference and diffraction. I'm sorry TRoc .. if we could just bear with spreading from the slit 'it happens' for a while.. (we have to start somewhere) Even a 'beam of light' is not understood. If you set up the ripple tank ( http://www.echalk.co.uk/Science/Physics/ri...Tank/ripple.htm ) for 'Setup Double Slit' then you see what are apparently 'beams' (of waves) propagating away from the slits. Also 'beams' of dark bits .. the beams of darkness are themselves an interference phenomenon .. take away the beam of brightness on one side of the dark line and you see that the remaining 'beam' spreads into that region because the 'other' wave that was producing the destructive interference has been removed. I'm set up with 'Setup Double Slit' 'Mouse=Edit Walls' I've drawn a wall at 45 degrees to reflect the central beam away to the right (starting about half way down the 'tank' ).. result as described above. There is a nasty sting in the tail of this analysis (for me) .. unless I can think of a way round it I have to claim that ALL beams are an interference effect Please play and report. Best wishes, -C2. This post has been edited by Confused2 on Feb 12 2007, 10:00 PM |
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| Aerohead |
Posted: Feb 12 2007, 10:04 PM
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Member ![]() ![]() Group: Members Posts: 55 Joined: 5-December 06 Positive Feedback: 80% Feedback Score: 6 |
Yes, I think you're right, C2. All waves diffract as a result of boundary conditions. The question now is: what are the boundary conditions through the slit ? It seems others want a model that accounts for it in terms of photon / electron interaction. And I think it no doubt is.
From the "for what it's worth department" of fluid mechanist's. ~Jim |
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| Confused2 |
Posted: Feb 13 2007, 01:08 AM
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Toothpaste salesman ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 4755 Joined: 8-November 05 Positive Feedback: 63.3% Feedback Score: -31 |
HI TRoc,
Excellent point. If there are two paths to start with then there are always two paths .. they do not form a 'beam' with nothing in the regions of destructive interference as Good Elf, Laserlight and Montec seem to be claiming. Can you convince them of that? Best wishes, -C2. |
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| Laserlight |
Posted: Feb 13 2007, 01:47 AM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
C2,
draw a 4 inch line down the middle of the dark bands. The idea is to put a center barrier between the bright bands. The dark band does not change, which it should if there is signal mixing taking place between the 2 bright bands on either side of the dark band. The signal remains the same even when the supposed wave mixing has a barrier in the way. The bands are just projections from the mixing point, or beams as you called them. Even without a screen the "beam" cones are "visible" in this model. What is your assessment after doing as I suggested? Regards, LL This post has been edited by Laserlight on Feb 13 2007, 01:53 AM |
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| Confused2 |
Posted: Feb 13 2007, 01:53 AM
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Toothpaste salesman ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 4755 Joined: 8-November 05 Positive Feedback: 63.3% Feedback Score: -31 |
Detection of anything that is there to be detected takes place at the point of detection. Always. Whether photon or EM .. it is the same .. because EM is photons. If there are two paths between the source and the point of detection then there are two paths between the source and the point of detection. Best wishes, -C2. This post has been edited by Confused2 on Feb 13 2007, 01:53 AM |
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| Laserlight |
Posted: Feb 13 2007, 02:08 AM
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Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 1158 Joined: 29-October 06 Positive Feedback: 100% Feedback Score: 8 |
C2 and TRoc, I drew an inverted funnel with the thin neck centered on the dark band. I got no secondary waveform interference. The center of the funnel stayed black. It appears that there is no energy propagating thru the center of the dark band. What results do you get doing the same set up. Regards, LL This post has been edited by Laserlight on Feb 13 2007, 02:15 AM |
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