## Joy's derivation of the 2*sqrt(2) bound

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

### Re: Joy's derivation of the 2*sqrt(2) bound

Rick Lockyer wrote:Fred,

The reason there is no equivalence is not because of some deficiency in GAViewer, It is doing the math correctly for both (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. The error is in Joy's position that [Ai , Bj] = 0, which is the basis for his simplification to 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. [Ai , Bj] = Ai Bj - Bj Ai which = 0 iff Ai = Bj by straight forward geometric algebra rules for multiplication.

So what? It can also equal zero iff Ai != Bj; (-1)*1 - 1*(-1) = 0. As usual, you don't understand the physics.

Rick Lockyer wrote:When I mentioned "triumph" I was being facetious. You captured Joy's sign error doing this, you did not accurately represent left and right handed bivector bases. I am truly amazed you were not able to pick this up from my derivation, especially after agreeing there were no geometric algebra errors in my presentation. Do the math with ei^ej right handed basis and ej^ei left handed basis and the extremely straight forward rules for reduction of geometric algebra wedge products instead of Joy's erroneous beta basis multiplication rules. I showed this, but do it for yourself. But really, it is actually as simple as (-1)(-1) = +1 = (+1)(+1). Think about it.

All you have to do is answer the simple question I asked you previously which you never answered.
viewtopic.php?f=6&t=183&start=110#p5108
FrediFizzx wrote:When looking at a left handed system from a right handed only frame of reference, is a x b in the left handed system going to be b x a in the right handed frame?

So what is your answer? And it is simply a yes or no answer.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

***
Lockyer continues his ignorant arguments, despite the fact that his errors have been repeatedly exposed elsewhere: viewtopic.php?f=6&t=183&start=90#p5085.

Rick Lockyer wrote:The reason there is no equivalence is not because of some deficiency in GAViewer, It is doing the math correctly for both (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. The error is in Joy's position that [Ai , Bj] = 0, which is the basis for his simplification to 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. [Ai , Bj] = Ai Bj - Bj Ai which = 0 iff Ai = Bj by straight forward geometric algebra rules for multiplication.

The GAviewer is not doing the physics correctly. The error Lockyer makes here is the same basic error he has been making for years. He thinks that one does physics by blindly applying mathematical rules. "That is not how physics works" is the first lessor any school teacher has to teach his/her students to liberate them from their ignorance. Some students quickly learn this lesson, but some never do. Lockyer never will. I would advice him not to give up his day job. In any case, as Fred already noted, the current physical problem is different from what Lockyer imagines. In Fred's demonstration (and in my analytical derivation) four separate experiments are put together in a single CHSH combination. Then the expression is squared before taking the square-root. This introduces second order effects in space-like separated events represented by the four A's and B's. In other words, in a given space-like region, only local non-commutativity should contribute to give the non-trivial result. This is an elementary lesson that any first year student of quantum field theory knows about, and it is also discussed in my papers linked by Fred. The end effect is that the unphysical contributions "occurring" at space-like separated locations are removed from the artificially squared factor before the square-root is taken. Any first year student of physics would understand this, but not our "mathematician." He requires spoon feeding: https://en.wikipedia.org/wiki/Tsirelson ... inequality.

In short, there is absolutely nothing wrong with Fred's demonstration above, or with my analytical derivation in the two papers linked by Fred in his original post.

Rick Lockyer wrote:When I mentioned "triumph" I was being facetious. You captured Joy's sign error doing this, you did not accurately represent left and right handed bivector bases. I am truly amazed you were not able to pick this up from my derivation, especially after agreeing there were no geometric algebra errors in my presentation. Do the math with ei^ej right handed basis and ej^ei left handed basis and the extremely straight forward rules for reduction of geometric algebra wedge products instead of Joy's erroneous beta basis multiplication rules. I showed this, but do it for yourself. But really, it is actually as simple as (-1)(-1) = +1 = (+1)(+1). Think about it.

As I already mentioned, no further comments needed here since Lockyer's errors have been thoroughly exposed elsewhere: viewtopic.php?f=6&t=183&start=80#p5045.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Rick Lockyer wrote:Fred,

The reason there is no equivalence is not because of some deficiency in GAViewer, It is doing the math correctly for both (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. The error is in Joy's position that [Ai , Bj] = 0, which is the basis for his simplification to 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2. [Ai , Bj] = Ai Bj - Bj Ai which = 0 iff Ai = Bj by straight forward geometric algebra rules for multiplication.

So what? It can also equal zero iff Ai != Bj; (-1)*1 - 1*(-1) = 0. As usual, you don't understand the physics.

Sure, real numbers are commutative for multiplication. But we are dealing with pure bivectors here. Did not think I had to explicitly cover this, I thought everyone was paying attention, even you.

FrediFizzx wrote:
Rick Lockyer wrote:When I mentioned "triumph" I was being facetious. You captured Joy's sign error doing this, you did not accurately represent left and right handed bivector bases. I am truly amazed you were not able to pick this up from my derivation, especially after agreeing there were no geometric algebra errors in my presentation. Do the math with ei^ej right handed basis and ej^ei left handed basis and the extremely straight forward rules for reduction of geometric algebra wedge products instead of Joy's erroneous beta basis multiplication rules. I showed this, but do it for yourself. But really, it is actually as simple as (-1)(-1) = +1 = (+1)(+1). Think about it.

All you have to do is answer the simple question I asked you previously which you never answered.
viewtopic.php?f=6&t=183&start=110#p5108
FrediFizzx wrote:When looking at a left handed system from a right handed only frame of reference, is a x b in the left handed system going to be b x a in the right handed frame?

So what is your answer? And it is simply a yes or no answer.

Yes it is simple: the coefficients change sign, which is all you appear to be looking at since you do not understand algebra. The rub is you cannot directly add right handed algebra coefficients to left handed algebra coefficients, they are attached to different basis elements, which in this case also have a sign change difference. Like I said above: " But really, it is actually as simple as (-1)(-1) = +1 = (+1)(+1). Think about it" Do you get it yet Fred?????

Both of you trot out the "you do not understand the physics" when I and others point out the math errors Joy has made and you fail to understand. Well when I was working on my physics degree at Stanford University, I was taught the physics is never in conflict with the math. If it appears to be, either one or both are incorrect or improperly applied. I do not know what algebra Joy is using if there are no errors in his claims. One thing for sure is it is not geometric algebra, for the errors are plain to see if it is.
Rick Lockyer

Posts: 46
Joined: Tue May 27, 2014 6:22 am
Location: Nipomo

### Re: Joy's derivation of the 2*sqrt(2) bound

Rick Lockyer wrote:
FrediFizzx wrote:So what? It can also equal zero iff Ai != Bj; (-1)*1 - 1*(-1) = 0. As usual, you don't understand the physics.

Sure, real numbers are commutative for multiplication. But we are dealing with pure bivectors here. Did not think I had to explicitly cover this, I thought everyone was paying attention, even you.

Ok, but the point here is that swapping the bivectors is a physics assumption and it looks to be an appropriate one since it produces the correct result.

FrediFizzx wrote:All you have to do is answer the simple question I asked you previously which you never answered.
viewtopic.php?f=6&t=183&start=110#p5108
FrediFizzx wrote:When looking at a left handed system from a right handed only frame of reference, is a x b in the left handed system going to be b x a in the right handed frame?

So what is your answer? And it is simply a yes or no answer.

Rick Lockyer wrote:Yes it is simple: the coefficients change sign, which is all you appear to be looking at since you do not understand algebra. The rub is you cannot directly add right handed algebra coefficients to left handed algebra coefficients, they are attached to different basis elements, which in this case also have a sign change difference.

So you answer "yes" but still don't see that by reversing the order of the GA products (a x b becomes b x a), we are converting the left handed system with the left handed coefficients to the right handed coefficients before summation. It's pretty simple. Not sure why you don't understand something that is so simple. I would have to say that it is you that doesn't understand the algebra.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Not sure why you don't understand something that is so simple. I would have to say that it is you that doesn't understand the algebra.

Fred, we have to simply accept that some people are just thick.

If they can't understand five lines of elementary algebra for so many years, even after our repeated explanations, then that is the only reasonable conclusion possible:
Joy Christian wrote:
With ${\beta_k(\lambda^i) = \lambda^i\,\beta_k}$ by construction (and $\lambda$ being the "hidden variable" in my model),

Here transition from Eq. (1.23) to Eq. (1.24) follows from geometric product, and transition from Eq. (1.25) to Eq. (1.26) follows from the fact that ${\lambda}$ is a fair coin.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Fred, I am not sure what you do not get about this:
Rick Lockyer wrote:The rub is you cannot directly add right handed algebra coefficients to left handed algebra coefficients, they are attached to different basis elements, which in this case also have a sign change difference.

You clearly do not understand the role/importance of basis elements for both addition and multiplication of n dimensional algebraic elements. Otherwise you would not be making these ignorant statements. Learn some algebra, you are clueless.
Rick Lockyer

Posts: 46
Joined: Tue May 27, 2014 6:22 am
Location: Nipomo

### Re: Joy's derivation of the 2*sqrt(2) bound

Rick Lockyer wrote:Fred, I am not sure what you do not get about this:
Rick Lockyer wrote:The rub is you cannot directly add right handed algebra coefficients to left handed algebra coefficients, they are attached to different basis elements, which in this case also have a sign change difference.

You clearly do not understand the role/importance of basis elements for both addition and multiplication of n dimensional algebraic elements. Otherwise you would not be making these ignorant statements. Learn some algebra, you are clueless.

The person who is utterly clueless here is Lockyer. He should learn some elementty algebra, some basic physics, and above all, how to read. For no one is adding right handed algebra coefficients to left handed algebra coefficients attached to different basis elements.

Fred, I urge you to stop responding to this extraordinarily thick troll. For other readers, I urge you to read my papers rather than falling for his idiotic comments:

Joy Christian wrote:I have presented my arguments in considerable detail using the correct basis elements routinely used in geometric algebra that Lockyer has either never bothered to read, or does not have the intellectual capacity to understand. This is in addition to the fact that he has absolutely no understanding of the physics involved in the EPR-Bohm type experiments, as the earlier comments by Michel Fodje in this thread made clear:

minkwe wrote:So you want Alice to map her results to Bob's basis in order to be able to do manipulate them together? Or you want Alice to map the result she gets for one particle, to the same basis as the result she got for the next particle in order to manipulate those outcomes? The outcomes are what they are, manipulating them as they are gives the correlations. Your arguments remind me of a proof that clockwise (+1) = counterclockwise(-1), because we are comparing readings of a single clock from two separate people Alice and Bob. And a third person Cindy, not privy to their bases, is calculating the correlation. A single person describing the clock can do so in two different bases, and both descriptions must agree with each other. In that case, you need a mapping between both. But once Alice and Bob are recording outcomes from their perspectives, and Cindy must calculate correlations from those outcomes without any knowledge of how the observation was made (cf HIDDEN VARIABLES), purists may indeed be so pedantic as to find that counterclockwise (+1) = clockwise (-1). Algebra does not help you in this situation without proper understanding of the physics that is going on.

As Paul Snively pointed out on his blog concerning the denial of Gill, Lockyer too is in denial of the computed demonstration of the correctness of my model.

As for anyone who wishes to verify, my detailed argument --- using the correct basis elements routinely used in geometric algebra --- can be found in the following papers, as I have repeatedly pointed out in this thread:

http://arxiv.org/abs/1211.0784

http://arxiv.org/abs/1203.2529

http://arxiv.org/abs/1501.03393

It is most unfortunate that I have had to deal with the childish tantrums of incompetent amatuers and liars like Gill, Moldoveanu, and Lockyer for so many years.

Fortunately there are well respected physicists and mathematicians who do understand my arguments perfectly well (in fact they are quite easy to understand):

Joy Christian wrote:In case anyone is wondering, the codes posted by Fred above inputs Eqs. (1) and (2) from the following one-page paper, and verifies Eqs. (5), (6), and (7):

Needless to say, any precocious schoolboy (apart from Gill, Moldoveanu, and Lockyer) can verify this set of equations analytically in less than 20 minutes.

A similar set of equations are also discussed in my book, as well as in this preprint, which is now published in the International Journal of Theoretical Physics:

These elementary set of equations are also verified, in great detail, by several prominent members of the foundations community, in particular by Lucien Hardy.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:

Although he is banned from this forum (just as he is banned from Paul Snively's blog), Richard Gill has "kindly" sent me the following admission in a private email:

Richard Gill wrote:
I am delighted to admit that I had overlooked the intended interpretation of your two equations: the interpretation which makes them entirely consistent with one another. In retrospect, very obvious ...

This was sent to me in a private email after he saw my following earlier post in this thread:

Joy Christian wrote:* * *

Well, Lockyer seems to have finally recognized his error. I hope he has the courage to admit his error in public (I expect no such decency from Gill and Moldoveanu).

Let me spell out the key point explicitly here so there remains no doubt in anyone's mind. The mathematical demonstration is exceedingly simple and easy to follow.

What we want to show is that the first equation above (as written by me in these two papers) forms a right-handed system; and the second equation (again, as written by me in these two papers) forms a left-handed system. Gill, Moldoveanu, and Lockyer falsely claim that I have made a sign error on the RHS of the second equation.

To check the handedness of these equations, let us set the angle between ${\bf a}$ and ${\bf b}$ to be 90 degrees and define ${\bf c} = {\bf a}\,\times\,{\bf b}$ for this case. Now multiply (using the geometric product) the first equation, on both sides, by ${(+{\it I}\,\cdot\,{\bf c})}$, from the left. Since we have set ${\bf a}$ and ${\bf b}$ to be orthogonal to each other, and since all bivectors such as ${(+{\it I}\,\cdot\,{\bf c})}$ square to ${-1}$, the first equation (in this special case of orthogonal ${\bf a}$ and ${\bf b}$) reduces to

${(+{\it I}\;\cdot\;{\bf a})\,(+{\it I}\,\cdot\,{\bf b})\,(+{\it I}\,\cdot\,{\bf c})\,=+1.}$

Now if we follow the same procedure for the second equation -- this time using the bivector ${(-{\it I}\,\cdot\,{\bf c})}$, it also reduces to

${(- {\it I}\;\cdot\;{\bf a})\,(-{\it I}\,\cdot\,{\bf b})\,(-{\it I}\,\cdot\,{\bf c})\,=+1.}$

So far so good. Both equations, as long as they remain "unaware" of each other, can be taken to represent a right-handed system, because their RHS equals to ${+1\,}$.

But now suppose we wish to compare the two systems, as done in the successive trials of EPR-B type experiments ( is the spin "up", or "down" in a given trial? ). Then we must find a functional mapping between the two equations. But that is a trivial task in the present set up, since we see that the respective bivectors are related as

${(-{\it I}\;\cdot\;{\bf n})\,=\,- \,(+{\it I}\,\cdot\,{\bf n}),}$

for any directional vector ${\bf n}$. If we now substitute the above mapping in the second of the two equations we have derived, we obtain at once, for this second system,

${(+{\it I}\,\cdot\,{\bf a})\,(+{\it I}\,\cdot\,{\bf b})\,(+{\it I}\,\cdot\,{\bf c})\,=-1.}$

Thus we see at once that the second equation represents a left-handed system with respect to the first system, because now the RHS of this equation equals to ${-1\,}$.

In conclusion, a sign mistake has indeed been made for the past several years, but it is made by Gill, Moldoveanu, and Lockyer. They should go back to their schools.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:Fred, I urge you to stop responding to this extraordinarily thick troll.

Yep, he is definitely trolling for sure now since his case got busted about a x b going to b x a upon conversion. Ok, I will stop if you stop.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Our goal in this post is to numerically verify Joy's derivation of the Tsirel'son bound $2\sqrt{2}$ from his original 2007 paper, and from his latest "Reply to Gill" paper.

We will use the GAviewer program for this purpose, as we did in our numerical verification of Joy's derivation of the EPR-B correlation from his one-page paper.

We begin by squaring the integrand of Eq. (18) in Joy's "Reply to Gill". We have relabelled the bivectors A's and B's to make them easier to check with GAViewer.

(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2)
= A1B1A1B1 + A1B2A1B1 + A2B1A1B1 - A2B2A1B1 + A1B1A1B2 + A1B2A1B2 + A2B1A1B2 - A2B2A1B2 + A1B1A2B1 + A1B2A2B1 + A2B1A2B1 - A2B2A2B1 - A1B1A2B2 - A1B2A2B2 - A2B1A2B2 + A2B2A2B2

Note that all of these bivectors are standardized variables, or standard scores, as explained by Joy in this post: viewtopic.php?f=6&t=196&start=40#p5426.

For example, A1 = +/-1 about a, B2 = +/-1 about b', etc.

Next, upon using the commutativity of A's and B's, [ Ai, Bj ] = 0 (because they are space-like separated), the terms in this expression can be collected as follows:

= A1A1B1B1 + A1A1B2B1 + A2A1B1B1 - A2A1B2B1 + A1A1B1B2 + A1A1B2B2 + A2A1B1B2 - A2A1B2B2 + A1A2B1B1 + A1A2B2B1 + A2A2B1B1 - A2A2B2B1 - A1A2B1B2 - A1A2B2B2 - A2A2B1B2 + A2A2B2B2

Note that neither A1 and A2, nor B1 and B2, commute, so their orders must not be altered.

Now, since all unit bivectors square to -1, the above expression further reduces to

= 1 - B2B1 - A2A1 - A2A1B2B1 - B1B2 + 1 + A2A1B1B2 + A2A1 - A1A2 + A1A2B2B1 +1 + B2B1 - A1A2B1B2 + A1A2 + B1B2 + 1

After some cancelations, this expression further reduces to

= 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2

= 4 + [ A1, A2 ] [ B2, B1 ] ,

as shown in Joy's two papers linked above. Either of these two results can be verified using the GAViewer, with the following code:

Code: Select all
//Adaptation of Albert Jan Wonnink's original code
//http://challengingbell.blogspot.com/2015/03/numerical-validation-of-vanishing-of.html
function getRandomLambda()
{
if( rand()>0.5) {return 1;} else {return -1;}
}

function crosspr(a, b)
{
c=a^b;
I=e1^e2^e3;
return c/I;
}

function getRandomUnitVector() //uniform random unit vector:
//http://mathworld.wolfram.com/SpherePointPicking.html
{
v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
return normalize(v);
}

batch test()
{
set_window_title("Test of Joy Christian's CHSH derivation");
N=20000; //number of iterations (trials)
I=e1^e2^e3;
s=0;
a1=getRandomUnitVector();
b1=getRandomUnitVector();
a2=getRandomUnitVector();
b2=getRandomUnitVector();

for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
{
lambda=getRandomLambda(); //lambda is a fair coin
//resulting in +1 or -1
mu=lambda * I;  //calculate the lambda dependent mu

A1=-mu.a1;
A2=-mu.a2;
B1=mu.b1;
B2=mu.b2;
q=0;
if(lambda==1) {q=(-(A2 A1 B2 B1)+(A2 A1 B1 B2)+(A1 A2 B2 B1)-(A1 A2 B1 B2));}
else {q=(-(B1 B2 A1 A2)+(B2 B1 A1 A2)+(B1 B2 A2 A1)-(B2 B1 A2 A1));}
s=s+q;
}
t=crosspr(a1, a2);
u=crosspr(b2, b1);
Scalar_Part=-4*t.u;
mean_F_A_B=s/N;
print(mean_F_A_B, "f");
print(Scalar_Part, "f");
prompt();
}

A typical result produced by the code, for example, is

mean_F_A_B = -0.079826 + 0.000216*e2^e3 + 0.000055*e3^e1 + -0.000286*e1^e2
Scalar_Part = -0.079826

So we can see that the scalar part of (- A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2) matches the factor -4*(a1 x a2).(b2 x a1) of Eq. (24) derived in Joy's paper. And the bivector part is vanishing.

In other words, we started with Eq. (18) of Joy's "Reply to Gill", and have now numerically verified Eq. (24) of that paper:

$\left|\,{\cal E}({\bf a}, {\bf b}) \,+\,{\cal E}({\bf a}, {\bf b'}) \,+{\cal E}({\bf a'}, {\bf b}) \,-\,{\cal E}({\bf a'}, {\bf b'}) \right|\,\leqslant\,\sqrt{4\,-\,4\,({\bf a}\times{\bf a'})\cdot({\bf b'}\times{\bf b})}$.

And since simple trigonometry dictates that

$-1\,\leqslant\,({\bf a}\times{\bf a'})\cdot({\bf b'}\times{\bf b})\,\leqslant\,+1,$

we finally see that after applying many iterations and taking the square-root gives us the bound $2 \sqrt{2}$ , because (as shown in the code) the bivector part vanishes.

To add to this post that numerically validates the analytical method of deriving $2 \sqrt{2}$, GAViewer will actually do the CHSH string calculation directly for Joy's model without any squaring of the integrand involved. Plus we will fix the vectors a1, a2, b1 and b2 to be nearly 2D like they would be in a typical experiment.
Code: Select all
//Adaptation of Albert Jan Wonnink's original code
//http://challengingbell.blogspot.com/2015/03/numerical-validation-of-vanishing-of.html
function getRandomLambda()
{
if( rand()>0.5) {return 1;} else {return -1;}
}

function getRandomUnitVector() //uniform random unit vector:
//http://mathworld.wolfram.com/SpherePointPicking.html
{
v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
return normalize(v);
}

batch test()
{
set_window_title("Test of Joy Christian's CHSH derivation");
N=20000; //number of iterations (trials)
I=e1^e2^e3;
s=0;
a1=1.00*e1 +0.01*e2 + 0.01*e3;
b1=0.707*e1 + 0.707*e2 + 0.01*e3;
a2=0.01*e1 + 1.00*e2 + 0.01*e3;
b2=0.707*e1 + -0.707*e2 + 0.01*e3;

for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
{
lambda=getRandomLambda(); //lambda is a fair coin,
//resulting in +1 or -1
mu=lambda * I;  //calculate the lambda dependent mu

A1=-mu.a1;
A2=-mu.a2;
B1=mu.b1;
B2=mu.b2;
q=0;
if(lambda==1) {q=(A1 B1)+(A1 B2)+(A2 B1)-(A2 B2);}
else {q=(B1 A1)+(B2 A1)+(B1 A2)-(B2 A2);}
s=s+q;
}
mean_F_A_B=s/N;
print(mean_F_A_B, "f");
prompt();
}

And the result is,

mean_F_A_B = 2.828200 + -0.000163*e2^e3 + -0.000069*e3^e1 !!!!!
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

Nice Fred!
Your success with the CHSH inspired me to create the full curve:

Code: Select all
function getRandomLambda()
{
if( rand()>0.5)
{
return 1;
}
else
{
return -1;
}
}

function getVector(idx)
{
step=0.1*pi; //step size is taken 1/10 of pi
angle=idx*step;
v=sin(angle)*e1+cos(angle)*e2+0.01*e3;
return v;
}
batch test()
{
set_window_title("Test Joy Christian");
N=20;
I=e1^e2^e3;
s[0]=0;
s[1]=0;
s[2]=0;
s[3]=0;
s[4]=0;
s[5]=0;
s[6]=0;
s[7]=0;
s[8]=0;
s[9]=0;
s[10]=0;
a[0]=0;
a[1]=0;
a[2]=0;
a[3]=0;
a[4]=0;
a[5]=0;
a[6]=0;
a[7]=0;
a[8]=0;
a[9]=0;
a[10]=0;
Ax[0]=
angleIx=0;
for(oo=0;oo<11;oo=oo+1) //iteration over aa
{
aa=getVector(oo);
for(pp=0;pp<11;pp=pp+1) //iteration over bb
{
bb=getVector(pp);
angleIx=oo-pp;
if(oo<pp)
{
angleIx=pp-oo;
}
minus_cos_a_b=-1*(aa.bb);

for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
{
lambda=getRandomLambda(); //lambda is a fair coin, resulting in +1 or -1

v=I.aa ;
w=I.bb;
q=0;
if(lambda==1)
{
q=v w;
}
else
{
q=w v;
}

s[angleIx]=s[angleIx]+q; //

a[angleIx]=a[angleIx]+1;

}
}
}
for(oo=0;oo<11;oo=oo+1) //iteration over angle differences
{
st=s[oo];
at=a[oo];
an=oo*10;
mean_mu_a_mu_b=st/at;
print(mean_mu_a_mu_b); //print the result
}

prompt();

}

which nicely results in:

mean_mu_a_mu_b = -1.00
mean_mu_a_mu_b = -0.95 + -0.00*e2^e3 + -0.00*e3^e1 + 0.01*e1^e2
mean_mu_a_mu_b = -0.81 + 0.00*e2^e3 + 0.00*e3^e1 + -0.03*e1^e2
mean_mu_a_mu_b = -0.59 + 0.00*e2^e3 + 0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = -0.31 + 0.00*e2^e3 + -0.00*e3^e1 + -0.05*e1^e2
mean_mu_a_mu_b = -0.00 + 0.00*e2^e3 + 0.00*e3^e1 + -0.07*e1^e2
mean_mu_a_mu_b = 0.31 + -0.00*e2^e3 + 0.00*e3^e1 + 0.04*e1^e2
mean_mu_a_mu_b = 0.59 + 0.00*e2^e3 + 0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = 0.81 + 0.00*e2^e3 + 0.00*e3^e1 + -0.04*e1^e2
mean_mu_a_mu_b = 0.95 + 0.00*e2^e3 + -0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = 1.00 + -0.00*e2^e3

(with each line the angle between a and b increases with 0.1 pi, starting from 0)

Last edited by ajw on Sat Sep 05, 2015 4:12 pm, edited 2 times in total.
ajw

Posts: 45
Joined: Sat Sep 05, 2015 2:04 pm

### Re: Joy's derivation of the 2*sqrt(2) bound

ajw wrote:which nicely results in (I have made an image of the resulting graph, but I see no way to upload it):

Hi Albert Jan,

Thanks. Email it to me and I will upload it and put it in your post.

Best,

Fred
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

ajw wrote:Nice Fred!
Your success with the CHSH inspired me to create the full curve:

Code: Select all
function getRandomLambda()
{
if( rand()>0.5)
{
return 1;
}
else
{
return -1;
}
}

function getVector(idx)
{
step=0.1*pi; //step size is taken 1/10 of pi
angle=idx*step;
v=sin(angle)*e1+cos(angle)*e2+0.01*e3;
return v;
}
batch test()
{
set_window_title("Test Joy Christian");
N=20;
I=e1^e2^e3;
s[0]=0;
s[1]=0;
s[2]=0;
s[3]=0;
s[4]=0;
s[5]=0;
s[6]=0;
s[7]=0;
s[8]=0;
s[9]=0;
s[10]=0;
a[0]=0;
a[1]=0;
a[2]=0;
a[3]=0;
a[4]=0;
a[5]=0;
a[6]=0;
a[7]=0;
a[8]=0;
a[9]=0;
a[10]=0;
Ax[0]=
angleIx=0;
for(oo=0;oo<11;oo=oo+1) //iteration over aa
{
aa=getVector(oo);
for(pp=0;pp<11;pp=pp+1) //iteration over bb
{
bb=getVector(pp);
angleIx=oo-pp;
if(oo<pp)
{
angleIx=pp-oo;
}
minus_cos_a_b=-1*(aa.bb);

for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
{
lambda=getRandomLambda(); //lambda is a fair coin, resulting in +1 or -1

v=I.aa ;
w=I.bb;
q=0;
if(lambda==1)
{
q=v w;
}
else
{
q=w v;
}

s[angleIx]=s[angleIx]+q; //

a[angleIx]=a[angleIx]+1;

}
}
}
for(oo=0;oo<11;oo=oo+1) //iteration over angle differences
{
st=s[oo];
at=a[oo];
an=oo*10;
mean_mu_a_mu_b=st/at;
print(mean_mu_a_mu_b); //print the result
}

prompt();

}

which nicely results in:

mean_mu_a_mu_b = -1.00
mean_mu_a_mu_b = -0.95 + -0.00*e2^e3 + -0.00*e3^e1 + 0.01*e1^e2
mean_mu_a_mu_b = -0.81 + 0.00*e2^e3 + 0.00*e3^e1 + -0.03*e1^e2
mean_mu_a_mu_b = -0.59 + 0.00*e2^e3 + 0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = -0.31 + 0.00*e2^e3 + -0.00*e3^e1 + -0.05*e1^e2
mean_mu_a_mu_b = -0.00 + 0.00*e2^e3 + 0.00*e3^e1 + -0.07*e1^e2
mean_mu_a_mu_b = 0.31 + -0.00*e2^e3 + 0.00*e3^e1 + 0.04*e1^e2
mean_mu_a_mu_b = 0.59 + 0.00*e2^e3 + 0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = 0.81 + 0.00*e2^e3 + 0.00*e3^e1 + -0.04*e1^e2
mean_mu_a_mu_b = 0.95 + 0.00*e2^e3 + -0.00*e3^e1 + -0.01*e1^e2
mean_mu_a_mu_b = 1.00 + -0.00*e2^e3

(with each line the angle between a and b increases with 0.1 pi, starting from 0)

That is really cool, Albert Jan. Going to take me a minute to fully understand the programming but this could be extended to even finer resolution. What is [oo]? Is that just part of the indexing?
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

oo is just an index that goes from 0 to 10. Your confusion might come from the fact that it is used twice: one time to iterate over the angles for aa, and one time to print the results (in a completely separated 4th loop).

So in the main part of the program the code picks an angle for aa based on the index oo, and uses this to create a 3d vector with the angle in the xy plane. Then, in an inner loop, it picks an angle for bb using the index pp.
In both cases the angles are calculated as: (indexAorB) * 0.1 * pi. That is why the difference between these indices (angleIx) can be used as a measure for the difference of the angle between aa and bb, and could be calculated also as (angleIx) * 0.1 * pi (not done in the current program).

To keep the angle difference positive, I check whether or not oo>pp.

So with angleIx being a measure for the angle difference between aa and bb, I can store all outcomes from the third inner loop having the same angle difference in the array s[angleIx]. Of course I also have to count how many times I have found a result for a certain angleIx, in a[angleIx]

The third inner loop just does the same as in the original code from my blog.
ajw

Posts: 45
Joined: Sat Sep 05, 2015 2:04 pm

### Re: Joy's derivation of the 2*sqrt(2) bound

***
Thank you, Fred and Albert Jan, for your excellent contributions.

By the way, the analytical counterpart of the latest simulations can be found in my "Reply to Critics" paper [cf. equations (1) to (5) on the first page of the paper].

***
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Taking a cue from Albert Jan's simulation we can do the a1, b1, a2, and b2 vectors more traditionally like this;

Code: Select all

a1=sin(0)*e1 + cos(0)*e2 + 0.0001*e3;
b1=sin(pi/4)*e1 + cos(pi/4)*e2 + 0.0001*e3;
a2=sin(pi/2)*e1 + cos(pi/2)*e2 + 0.0001*e3;
b2=sin(7*pi/4)*e1 + cos(7*pi/4)*e2 + 0.0001*e3;

So we can see that the Bell test angles in degrees are a1 = 0, b1 = 45, a2 = 90, and b2 = -45. And the result for the CHSH GAViewer calculation for Joy's model is;

Joy_CHSH = 2.828427 + -0.000000*e2^e3 + -0.000000*e3^e1

My calculator shows that $2 \sqrt{2} = 2.828427$.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Taking a cue from Albert Jan's simulation we can do the a1, b1, a2, and b2 vectors more traditionally like this;

Code: Select all

a1=sin(0)*e1 + cos(0)*e2 + 0.0001*e3;
b1=sin(pi/4)*e1 + cos(pi/4)*e2 + 0.0001*e3;
a2=sin(pi/2)*e1 + cos(pi/2)*e2 + 0.0001*e3;
b2=sin(7*pi/4)*e1 + cos(7*pi/4)*e2 + 0.0001*e3;

So we can see that the Bell test angles in degrees are a1 = 0, b1 = 45, a2 = 90, and b2 = -45. And the result for the CHSH GAViewer calculation for Joy's model is;

Joy_CHSH = 2.828427 + -0.000000*e2^e3 + -0.000000*e3^e1

My calculator shows that $2 \sqrt{2} = 2.828427$.

Those four vectors are actually just a tiny tiny bit longer than unity because we have 0.0001*e3 for the z direction. But it doesn't matter much. And if we set the e3 coefficients to zero for all four vectors we get;

Joy_CHSH = 2.828427

No bivector residue at all.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Previous