DrChinese said:
1. In EPR, we use one entangled particle (say the Right) to gain knowledge of the other (say the Left \theta). A measurement of the Right tells us it is polarized at 0 or 90 degrees relative to an arbitrary angle we shall call 0 degrees.
Sorry, but I'm afraid you've already gone wrong! A single measurement of polarisation does not, in classical theory, tell you what the input polarisation was.
The classical formula, given that the input polarisation was in a random direction, is:
P_{AB}(a, b) = <br />
\int^{\pi}_0 \frac{d\lambda}{\pi} \cos^2(\lambda - a)\cos^2(\lambda - b)
which evaluates to
\frac{1}{8} + \frac{1}{4} \cos^2 \theta
where \theta = b - a.
The logic behind this is well known, even to quantum theorists, and reproduced in many papers, including the one I keep referencing,
http://arXiv.org/abs/quant-ph/9903066.
... Assuming the light is randomly polarized initially so the Right side odds are 50-50, we get:
P(correlation)=P(+,0)+P(-,90)=.5cos^2(\theta)+.5sin^(90-\theta)=cos^2(\theta)
which is also the QM prediction.
No, this is not how to deal with random polarisation directions in classical optics. It's some kind of compromise, influenced by QM thinking and the photon idea. In classical optics the initial beam has a definite polarisation direction, \lambda, i.e. it has the very "hidden variable" assumed in all (reasonable) local realist models for optical Bell tests. This is commonly known, reproduced in many texts other than mine.
If you care to make a case that this is not how the rest of the world looks at this problem, by all means, go ahead. To me, it looks like dismissive hand-waving on your part.
I'm afraid that, to me, you algebra looks like a quantum-theoretical fudge!
I have seen your charts on possible values per LR ...
But I have never given any! If you are using anything out of my Chaotic Ball papers then you are using it incorrectly. As I've said before, it illustrates the principle only. As I've also said before (and as you might now better understand, with the aid of the equation above), all that is needed to convert the basic, idealised, local realist (optical) Bell test prediction into one that can have higher "visibility" is to alter the assumed shapes of the functions involved, allowing for the true operating characteristics of the apparatus concerned. There is no need for the model to predict values anywhere near as high as 0.75. The high figure you quote is after "normalisation".
(As to your idea that the LR model predicts both + and - cases simultaneously... hey, you are the one that is talking to us about what is REASONABLE and INTUITIVE. And that doesn't sound so reasonable to me. In fact, it sounds like QM weirdness if you think about it

)
Hmmm ... Well now, possibly (after seeing the equation) you will by now see for yourself why both + and - is perfectly natural, but just in case you don't, consider what happens if you input light polarised at 45 deg to the axes of a 2-channel polariser.
Suppose the initial intensity is I. The intensity in
both outputs will then (under the usual assumption of a 50-50 split) be I/2. Assuming "perfect" detectors, giving a count every time for a signal of intensity I and for signals of lesser intensity a count with probability proportional to that intensity, the detector for the + channel has a probability 0.5 of a count and likewise, independently, the - channel. The probability of both at once is 0.25.
Is that so very weird?
You will have noticed, incidentally, that the above depends on a number of assumptions, none of which are absolutely necessary. You can get different answers under different conditions. As has been shown experimentally, when you have very weak light (in QM considered to be "single photon" level), the individual pulses making up the beam may not split 50-50. That, though, is a separate issue, not critical here.
Caroline
http://freespace.virgin.net/ch.thompson1/