How to generate correlations of ANY strength: Linear to Box

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

Re: How to generate correlations of ANY strength: Linear to

Postby FrediFizzx » Wed Mar 12, 2014 11:27 am

Ok, you all. Thread has gone off topic. Let's get back to the main topic of this thread or I will lock it.
FrediFizzx
Independent Physics Researcher
 
Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Wed Mar 12, 2014 11:39 am

FrediFizzx wrote:Ok, you all. Thread has gone off topic. Let's get back to the main topic of this thread or I will lock it.


Sorry, Fred. Let me try to get back to the main topic.

A better way of stating what I have stated in the first post of this thread is that the ratio can be varied to produce correlation of any strength, from Linear to Box:


Linear: Box.


This further supports my view that quantum correlations are purely geometrical and topological effects. They have nothing to do with non-locality or non-reality.

Now I find this observation very interesting, but so far it has not attracted much interest.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Wed Mar 12, 2014 11:59 pm

The above ratios are for the "initial" function .
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Fri Mar 14, 2014 2:49 am

Joy Christian wrote:Now I find this observation very interesting, but so far it has not attracted much interest.

Actually it is well known among those who study the detection loophole, that one essentially can get any correlation one likes from classical *through quantum* all the way to the Rohrlic-Popescu local boxes. In other words, the Tsirelson bound is no bound anymore. Many people have remarked that the really interesting things in the real world experiments like Aspect's, Weihs', and so on, is that we never appear to violate the Tsirelson bound 2 sqrt 2.

Actually a colleague of mine in Leiden, a well known, senior, respected experimenter in quantum optics (e.g.: member of the Royal Dutch Academy of Sciences), did recently publish an experiment with experimental violation of the Tsirelson bound. He thought he had established stronger forms of entanglement than previously observed, hence his experiment was better than earlier ones, when in fact, taken at face value, he had disproved quantum mechanics. Amusing that such a paper gets published in one of the best journals, and no one notices that there is something odd about the result.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Fri Mar 14, 2014 5:26 am

gill1109 wrote:...the really interesting things in the real world experiments like Aspect's, Weihs', and so on, is that we never appear to violate the Tsirelson bound 2 sqrt 2.


The reasons for this fact have been explained by me in exquisit details in this paper: http://arxiv.org/abs/1101.1958.

Let me also stress once again that my model has nothing whatsoever to do with detection loophole, or any other loophole for that matter.

I have shown in the above paper why, not only the EPR-Bohm correlation, but also no quantum correlation can exceed the strength of .

The true reason for the existence of this bound has to do with the torsion in our physical space on the one hand, and the parallelizability of the four spheres corresponding to the four normed division algebras on the other hand. More information about this can be found on my blog: http://libertesphilosophica.info/blog/

The observation of the Linear to Box correlation I made in my previous posts is interesting to me in the light of these results. In particular, it is interesting from my perspective that the exact geometrical ratio , and only the exact geometrical ratio , is responsible for the quantum correlations. This further supports my view that they have nothing to do with non-locality or non-reality. Quantum correlations are simply correlations among the points of the non-trivially parallelized spheres.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sat Mar 15, 2014 2:35 am

Joy Christian wrote:Let me also stress once again that my model has nothing whatsoever to do with detection loophole, or any other loophole for that matter.

Amusing that your computer simulation of your model is "lifted" straight from publications about the detection loophole.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Sat Mar 15, 2014 3:40 am

gill1109 wrote:
Joy Christian wrote:Let me also stress once again that my model has nothing whatsoever to do with detection loophole, or any other loophole for that matter.

Amusing that your computer simulation of your model is "lifted" straight from publications about the detection loophole.


Amusing that at other junctures you have stressed quite stressfully that one and the same mathematical structure can describe many different physical phenomena.

Two apples and two bananas have the number 2 in common. Therefore apples must be bananas?

Those who are more serious and less cynical about my work may find the supposed "coincidence" not so surprising, especially after reading this paper: http://libertesphilosophica.info/blog/w ... hapter.pdf. After all, both publications are trying to explain the same number , namely the EPR-Bohm correlation.

What is more, as I have stressed many times before, a simulation of an analytical model is not the model itself. It is merely a simulation of the model.

Finally, the initial or complete state in the publications about detection loophole is a lonesome vector ,

whereas the initial or complete state (in one of the representations of my 3-sphere model) is a pair .

Apples versus oranges.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sat Mar 15, 2014 4:15 am

Joy Christian wrote:Amusing that at other junctures you have stressed quite stressfully that one and the same mathematical structure can describe many different physical phenomena.

Touché!
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Heinera » Mon Mar 17, 2014 6:12 am

gill1109 wrote:PS Florin Moldoveanu wrote me the following (note: I disagree with his correction of equation 22)

I double checked Pearl's computations, and I found many problems. All checks out until formula 16 (including 16 and including appendix A and B) Then it goes downhill. (17) has a typo, the correct formula is:

mu(x) = - 1/(1-x^2)^{1/2} d/dx [(1-x^2)^{1/2} / 4x d/dx g(x) x^2]

Equation (19) is messed up big time, 1/2 g(x) is the sum of the 2 integrals over [(1-x^2)^{1/2} + x] and not the first integral over (1-x^2) + the second integral over x^2. As a consequence Eq. 23 is wrong as well, but 23 does follow from the incorrect 19. Section V of the paper is based on (23) and is invalid.

I could not obtain Eq. 20 (I got a similar slightly more complex one-I am still working on it and maybe it will simplify to 20)

Eq. 21 has a typo and should read:

h(x) = 1/4 C pi 1/(1+x)^3

Eq. 22 has a typo and should read:

rho(r) r^2 = 1/3 sin (1/2 pi r) / (1+cos(1/2 pi r))^3

It was a major pain to straighten out the math on page 1421 because it is a sloppy paper and there are many false trails one need to explore to understand the intention. On Eq. 20 I got a very similar expression but I am lacking the overall 1/1-x^2 factor outside the square brackets, and in the second integral I have an additional term of (1-z^2)^{1/2}. The first integral in the square bracket is equal with g(0)/2.


I think Florin is actually right about Eq. 22 here. If we take Richard's cdf integral

4/3 . ( -1 / 4 + 1 / (1 + cos pi r / 2) ^2 )

and check by differentiating it back to get the distribution, we get Florin's Eq. multiplied by 4 pi.
Heinera
 
Posts: 917
Joined: Thu Feb 06, 2014 1:50 am

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Mon Mar 17, 2014 6:21 am

Heinera wrote:
gill1109 wrote:PS Florin Moldoveanu wrote me the following (note: I disagree with his correction of equation 22)

I double checked Pearl's computations, and I found many problems. All checks out until formula 16 (including 16 and including appendix A and B) Then it goes downhill. (17) has a typo, the correct formula is:

mu(x) = - 1/(1-x^2)^{1/2} d/dx [(1-x^2)^{1/2} / 4x d/dx g(x) x^2]

Equation (19) is messed up big time, 1/2 g(x) is the sum of the 2 integrals over [(1-x^2)^{1/2} + x] and not the first integral over (1-x^2) + the second integral over x^2. As a consequence Eq. 23 is wrong as well, but 23 does follow from the incorrect 19. Section V of the paper is based on (23) and is invalid.

I could not obtain Eq. 20 (I got a similar slightly more complex one-I am still working on it and maybe it will simplify to 20)

Eq. 21 has a typo and should read:

h(x) = 1/4 C pi 1/(1+x)^3

Eq. 22 has a typo and should read:

rho(r) r^2 = 1/3 sin (1/2 pi r) / (1+cos(1/2 pi r))^3

It was a major pain to straighten out the math on page 1421 because it is a sloppy paper and there are many false trails one need to explore to understand the intention. On Eq. 20 I got a very similar expression but I am lacking the overall 1/1-x^2 factor outside the square brackets, and in the second integral I have an additional term of (1-z^2)^{1/2}. The first integral in the square bracket is equal with g(0)/2.


I think Florin is actually right about Eq. 22 here. If we take Richard's cdf integral

4/3 . ( -1 / 4 + 1 / (1 + cos pi r / 2) ^2 )

and check by differentiating it back to get the distribution, we get Florin's Eq. multiplied by 4 pi.

Yes you're right and I was wrong. I had forgotten a factor pi/2 from the transformation from r in [0, 1] to an angle theta = r pi/2 in [0, pi/2]
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Previous

Return to Sci.Physics.Foundations

Who is online

Users browsing this forum: ahrefs [Bot] and 70 guests

cron
CodeCogs - An Open Source Scientific Library