Lazarus

Your last reply persuaded me to post a very simple explanation of why Bell’s assumption that doubling the angle doubles the wrong way photons has a problem. The rectangular measurement devices are 2.5 meters long and 2 meters wide and both of the measurement devices are rotated the same amount. . The polarized photons are 2.0001 meters long. When the photons are aligned parallel or perpendicular 100% are opposite. Successive photons have a random location spread of .1 meters. When one of the measuring devices is rotated 1% the length available to a wrong way photon increases to 2.000304656 meters. That allows some of the wrong way photon to get through. Rotating the measuring device 2 degrees makes the available length 2.001219088 meters. The slack allowing some of the wrong way photons to get through is .000304656 meters for a 1% rotation and .001219088 meters for a 2% rotation. The slack is not double but close to 4 times which implies that 4 times as many wrong way photons can get through. So it is NOT impossible to conceive of a correct physical interpretation.

You see black, I see white. I greatly respect your knowledge, intellect, experience and willingness to share your knowledge. In this case I just cannot agree with your perspective. If I can come up with an analysis that I think would be more acceptable to you I will post it. You have always given helpful responses to any post and I am grateful for your help.

Regardless of whether or not the Bowling Ball model is a valid demonstration of the failure of the assumption that doubling the angle of rotation of a measuring device can’t more than double the misses, what justifies Bell’s assumption that it is impossible for a physical situation to more than double the number of misses.

The Bell Inequality proof makes the unwarranted assumption that rotating the measuring device two degrees would necessarily limit the number of misses to twice the number of misses of a one degree rotation. That assumption is not only unjustified but contrary to experimental evidence. The Bowling Pin description demonstrates one example that does not fit Bell’s proof. An interesting comment by Bell was made. Later in his life, Bell expressed his hope that such work would "continue to inspire those who suspect that what is proved by the impossibility proofs is lack of imagination”.

The length of the pin is less than the long dimention of the rectangle and greater than the width of it. With the rectangle in the long derecttion vertical, a vertical pin will always go through. A horizontal pin will always hit the side of the rectangle. When the rectangle is rotated a bit the horizontal distance becomes greater and a hoizontal pin has a better chance of making it through without hitting the side.

The following appears to be an explanation of the “proof” of quantum entanglement. Imagine a pair of particles that can be measured at distant locations. Suppose that the measurement devices have settings, which are angles—e.g., the devices measure something called spin in some direction. The experimenter chooses the directions, one for each particle, separately. Suppose the measurement outcome is binary (e.g., spin up, spin down). Suppose the two particles are perfectly anti-correlated—in the sense that whenever both measured in the same direction, one gets identically opposite outcomes, when both measured in opposite directions they always give the same outcome. The only way to imagine how this works is that both particles leave their common source with, somehow, the outcomes they will deliver when measured in any possible direction. (How else could particle 1 know how to deliver the same answer as particle 2 when measured in the same direction? They don't know in advance how they are going to be measured...). The measurement on particle 2 (after switching its sign) can be thought of as telling us what the same measurement on particle 1 would have given. Start with one setting exactly opposite to the other. All the pairs of particles give the same outcome (each pair is either both spin up or both spin down). Now shift Alice's setting by one degree relative to Bob's. They are now one degree off being exactly opposite to one another. A small fraction of the pairs, say f, now give different outcomes. If instead we had left Alice's setting unchanged but shifted Bob's by one degree (in the opposite direction), then again a fraction f of the pairs of particles turns out to give different outcomes. Finally consider what happens when both shifts are implemented at the same time: the two settings are now exactly two degrees away from being opposite to one another. By the mismatch argument, the chance of a mismatch at two degrees can't be more than twice the chance of a mismatch at one degree: it cannot be more than 2f. Compare this with the predictions from quantum mechanics for the singlet state. For a small angle θ, measured in radians, the chance of a different outcome is approximately as explained by small-angle approximation. At two times this small angle, the chance of a mismatch is therefore about 4 times larger, since . But we just argued that it cannot be more than 2 times as large. This intuitive formulation is due to David Mermin. The small-angle limit is discussed in Bell's original article, and therefore goes right back to the origin of the Bell inequalities. ------------------------------------------------------- There are a couple of things that need a better explanation. The contention that the results for 2% implies twice as many misses has a problem. As shown by the bowling pins, Rotating the target area does not linearly increase the length of a horizontal line across the target area. That means the two times as many assumption has a problem. The other item is that it doesn’t seem to “prove” the flight of the photon stayed indeterminate. A slight rotation of the target area allows some misses but doubling the rotation can more than double the misses.

The state of the bowling pins is undetermined while the firing mechanism is rotating. When the pins are fired the state is no longer undetermined but is unknown. How do we know the state of photons remains undetermined while they are in flight?

When you measure the state of one object, why would anything need to be transmitted to the other since it has been in the same state and relationship since it became entangled?

If you know the relationship of 2 objects and the state of one of them doesn’t that allow you to calculate the state of the other?

This effort was an attempt to match the results of the common explanations of entanglement results. There is something missing in most explanations. If the state of the entangled objects is the same at the start and doesn’t change in flight, it should give the same results at any time the objects are measured. Two entangled photons with the same polarization and frequency should be the same at any time that they were examined. What is missing that causes the problem?

Swansont said: No it can't if it works as you describe, measuring pairs. If the feed is random you can get uu, dd and ud pairs. Lazarus said: The randomness is between paths. Two bowling pins are released at a time from one path. Only uu or dd. ---------------------------------------------------------------------------------------------------------------------------------------------- Swansont said: Further, what happens if you change the detection axis? What if it's at 45° to the device axis? 90°? Lazarus said: Excellent point. To make it match better the bowling pins must be magnetized with North on the small end and South on the big end. The measuring devices are each a pair of magnetism sensors on both sides of the target areas. The two sensors connect to a decision box that generates a one or zero based on which input signal is stronger. The target area is a little larger than the bowing pins. At 0 and 180 degrees the results are 100%. At 90 degrees the results are 50% correct. In between the percentage varies.

The selection of which side feeds into the projection mechanism can be may as random and unpredictable as desired. How does that make the initiation determined?

The up pins stay up, the down pins stay down. To the observer that is going to measure them they are undetermined before measurement.

What you are getting at is that the paired things are suppose to oscillate and the bowling pins do not. Is the evidence for oscillation that solid? For instance, photons are represented as sine waves but that may not be the physical case. The input to an antenna is designed to be a sign wave so the output appears as a sign wave. But since a photon is apparently generated by a single electron and will cause an electron in an appropriate wire to move in one direction a second photon is needed to move an electron in the opposite direction. It looks like a sine wave to the receiving antenna. Other effects could be causing the appearance of oscillation in many cases.

Same rules

UNGHOSTLY ENTANGLEMENT A method of entangling pairs of bowling pins. Two lines of bowling pins feed into a device that shoots them off two pins at a time. One line of pins is upright and the other line has the pins inverted. Each line feeds two pins at a time into the shooting device. The two lines of pins feed alternately or randomly into the shooting device. Measuring the results at random times will have 50% up pins and 50% down pins. However, if one pair of pins is fired then measured both pins will be either up or down, regardless of the time or place of the measurement.

Swansont said: Do you have evidence to back up your claim? Do you have a calculation that predicts the size of the interference pattern? If yes, please present them. Strange said: Or how about: you show us your calculations? ------------------------------------------------------------------------------------------------------------------------------------------------------------------ Here is a simulation of Diffraction of Baseballs. The numbers are the number of times a baseball landed in that square. SINGLE EDGE 1232211221 233223133 2 11 32313 211222 35332111 1 33444452 1 1 1444233312 13222211221 12321 1353322322 254333133 2 12111 1132313 4223331 1 35333212 2 33444453 2 2 1655344312 24333211221 12321 2464422432 35433423412 1322211222213 1534344211 46443212 2 33443444 3 3 1655455312 36544423322 1331 2464422442 35333322221 2233222312 3 1534344211 46443212 2 3344233413 4 1655455312 34222322311 1331 2464422432 25544312322 22333333 2 2 1534233211 35333212 2 1433312332314 5434442 2 242222122 1 12321 1353322322 14433312322 22334342 1 1 1323122211 35332111 1 13222 2242414 3323331 1 221112122 1 12321 1232211221 14433211221 22334341 1323122211 23222111 1 12111 1142414 * 211222 221112122 1 12321 DOUBLE SLIT 12323344421221 2332464553333 2 11 11 323162313 2112431222 353356442211 1 334477964553 1 1 14443777353312 132235334421221 123222321 13534576542322 2543585663333 2 12112222424262313 42237533431 1 353367453412 2 334477974655 2 2 16554998464312 243356445421221 123222321 24646688742432 354369667543412 13223434433352213 15344976454211 464478563412 2 334467883747 3 3 16555998575312 365479777643322 13311331 24646688842442 353368555532221 22334456342612 3 15344976454211 464478563412 2 33445678363813 4 16555998575312 342257445432311 13311331 24646688742432 255468667532322 223355663535 2 2 15343866343211 353367453412 2 14334556635472314 54349854642 2 2422463442222 1 123222321 13534576542322 144347556532322 223365754443 1 1 13232545232211 353356442211 1 13223344626382414 33236633431 1 2211342332222 1 123222321 12323344421221 144346545421221 223365744341 13232545232211 232244332211 1 12112222525282414 * 2112431222 2211342332222 1 123222321 Y Single Edge Setup . . ........................................................................ . . . . b . u Floor . m-----------------------------------------------------------------X p . s . . . . . .......................................................................... . . Even with this crude simulation the diffraction patterns are evident. Also, this explains why the double splits have a stronger diffraction pattern than the Single Slit result. The overlapping patterns make the difference between the highs and lows greater. The mechanical set up is a Single Edge or a Double Slit, 39 feet above the floor with counters for each square foot of a 21 foot by 79 foot rectangle. The uneven edge (bumps) consists of 1 foot squares The Double slits are 3 feet wide, including the “bumps”. The space between the slits is 1 foot. The “bumps” are sloped down from horizontal at 3 different angles, 30, 39 and 51 degrees.. The “bumps” are bent down in the forward direction, then the sides are bent down with the horizontal size still 1 foot square.” Since the Double Slits are equivalent to 4 Single Edges all that is needed for the Double Slit version is to overlay one pattern with another shifted by the distance from one slit to the other which is 4 feet. One hundred baseballs were dropped on each “bump”, evenly spaced. The calculations The resulting angle of the bounce, A = s*(h^.5*f + 1)*(h*g + 1) The horizontal direction vector of the bounce = (1,(j-i)*0.5) The horizontal distance of the bounce = (39*cos(A)) / sin(A) Where A is the angle from horizontal of the bounce. s is the angle from horizontal of the bump before bending. (30, 39 or 51 degrees) f is the bend factor to bend the x direction of the bump down. (0.17) g is the bend factor to bend the sides of the bump down. (0.1) h is the distance along the x axis of the hit of the baseball on the bump. i is the distance along the y axis of the hit of the baseball on the bump. j is the location of the left center of the bump on the y axis. A better simulation program could produce a pattern to match light diffraction as closely as desired but the basic concepts are the same.

No real surface is perfectly smooth. I will try to produce something to show that a reasonable diffraction pattern can be created.

When you produce a pattern that matches a light Dpuble Slit pattern, you can use that to speculate on the shape of the surface that could cause the photons to react similarly. Everything that I have read clains that the reason the Young Double Slit Experiment implys a wave is that it is impossible for a partical to produce the pattern. Wasn't that the original argument.?

Strange said: I think he is trying to show that you don't need quantum effects to generate an interference pattern using solid objects. By contriving a set of suitably sized and shaped objects for the balls to bounce off, he is (I think) claiming that the balls will land in a pattern resembling a diffraction pattern. (The word quincunx comes to mind: http://www.mathsisfu.../quincunx.html) And therefore (look out for the leap in logic) light is not waves! ----------------------------------------------------------------------------------------- Reply: Creating a pattern similar to the light defraction is what it is trying to do. Adjusting the shape and number of pyramids with adjusting the sizes almost any pattern could be produced. Producing a similar pattern does not say that a phonon is not a wave, merely that there are other possibilites. Thank all of you for your comments. Lazarus

Did you take into account that the pyramids are 3 different hights so the angle of bounce is different for each type?

DOUBLE SLIT EXPERIMENT WITH BASEBALLS First we do the experiment for single edge diffraction Make a horizontal edge 3 feet wide with a thickness of 6 inches. Then cover the 6 inch surface with 4 sided pyramids with their square bases 6 inches wide. Use pyramids with 3 different heights. 12 inches, 9 inches and 6 inches. When we drop baseballs from 10 feet above the baseballs will land in one of 3 lines on the floor as they bounce off of the pyramids. Now round off the exposed edges of the pyramids. That causes the baseballs to land in 3 wider lines to form the diffraction pattern. The next thing to do is build a double slit replica with rounded pyramids on all the 6 inch surfaces. Now the baseballs will land in multiple fuzzy distinct lines similar to Young’s Experiment. If this makes sense to you, the Double Slit Experiment does not offer strong evidence that photons are waves. Electrons in elliptical orbits in atoms could make the leading edge of a razor blade bumpy to produce the diffraction effect. I have previously shown that a rotating magnetic force in the nucleus of atom would account for the discrete orbits of electrons. In single edge diffraction photons do not go back under the edge. That makes the pinhole diffraction interference explanation suspicious.

What is the limit of (1-1/n)^n? For (1+1/n)^n it is e, about 2.7 but I can't find the negative result.

Swansont said: I just ran the numbers, and using the critical frequency from http://en.wikipedia....otron_radiationfor hydrogen (assuming a Bohr orbit) the wavelength is 3.5 x 10-11 m. i.e. smaller than the atom. (It also corresponds to a 35 keV photon, which is a pretty hefty x-ray.) Reply: Yes, an electron with that energy would blaze its way out of an atom. In a static atom the energy levels do not reach that high. The escape energy (kinetic energy) of the lowest level of hydrogen is 13.6 eV or a wave length of about 9*10^-8 meters. More importantly, the escape energy for a uranium nucleus with only one electron is about 12,500 eV or 10^-10 meters compared to the size of the uranium atom of about 3*10^-10 meters. So you are right that at the extreme end of the periodic chart the wave length could be smaller. The normal configuration is to have 2 electrons in the lowest level which would mean an escape energy of about 3800 eV or 3*10^-10 which comes closer. It might well be that uranium atom with only one electron would be unstable. xyzt said: Electron does not "approach from the center". You cannot make up sh!t and throw it around trying to see if it sticks to the wall. Reply: Perhaps you should read the description of the hypothetical experiment again.

Swansont said: The radiation in a synchrotron is not toward the center Reply: If an electron approaching from the center changed direction by 180 degrees the radiation would go to the center. Swansont said: How do you know the photon would be much larger than the atom? Reply: The diameter of atoms is about 10^-10 meters. The wave length of visible light is about 10^6 meters. Soft X-Rays are about 10^9 meters. Since a polarizing grate with spacing less than a wave length will block some photons at least one direction across the photon must approximate the wave length. Swansont said: Why would you expect the energy from an atomic electron to go into the nucleus (if it worked this way, which it doesn't)? Reply: A ball bouncing off a wall slows down as it transfers energy to the wall. Similarly, an electron “bouncing” off a synchrotron or an electron making one half of a cycle around a nucleus must transfer energy and slow down. In the synchrotron it appears that there are 2 transfers of energy away from the electron which should be offset by the slowing of the electron. One is the radiation and the other is the reaction to the “bounce” of the electron. One possible solution is that the electron slows enough to compensate for both.