## 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

FrediFizzx wrote:
Joy Christian wrote:These are all commuting numbers, so mathematically this is quite a different argument.

Is there any way to tell the GAviewer that A's and B's commute for the first F?

Well, the point is that it is possible to show a violation with "dependent" terms. It is a subtle thing on exactly what the dependency is.

I don't think GAViewer will do that but I will investigate it further.

I don't think this is right. Consider the following: Let $f(u)$ be a function of a single variable, any function, defined such that it has range

$-1 \leq f(u) \leq 1$

Let us for the moment not care about the domain of the function. But it immediately follows that for 4 independent variables in it's domain $w, x, y, z$, any linear combination of the function over those variables is bounded above by 4 and below by -4:

$-4 \leq f(x) \pm f(y) \pm f(z) \pm f(w) \leq 4$

It turns out that the calculation often performed by experimentalists using experimental expectations, and also that predicted by using quantum mechanics is similar. It is

$f(x) + f(y) + f(z) - f(w)$

Since we already know that those terms relate to statistically independent measurements, we should conclude that the upper bound is 4 as well. However, the variables $w, x, y, z$ in the EPRB scenario are not completely independent. Often the expression is written as:

$f(a-b) + f(a-b') + f(a'-b) - f(a'-b')$
which is exactly equivalent to
$f(x) + f(y) + f(z) - f(y + z - x)$

Therefore we have 3 independent variables only. What then is the upper bound for this expression. With a little algebra, we find that the upper bound is also the maximum of the expression

$3f(x) - f(3x)$

Which corresponds to 2 sqrt 2 when f(x) = cos(x). (http://www.wolframalpha.com/input/?i=Ma ... os%283x%29)

On the flip side, if we know that the terms are counterfactual as in the CHSH or Bell's inequality, then there is more dependence between the terms than just the settings as explained above, and the upper bound is 2 because

$f(x) = g_i(a)g_i(b)$
$f(y) = g_i(a)g_i(b')$
$f(z) = g_i(a')g_i(b)$
$f(w) = g_i(a')g_i(b')$

where $-1 \leq g_v(\cdot) \leq 1$
therefore the expression reduces to

$g_i(a)g_i(b) + g_i(a)g_i(b') + g_i(a')g_i(b) - g_i(a')g_i(b')$

Which factors as

$g_i(a)[g_i(b) + g_i(b')] + g_i(a')[g_i(b) - g_i(b')]$
And obviously the upper bound for this expression is 2. This is the CHSH. However, this analysis does not work for experiments and QM because instead of exactly the same functions $g_i(\cdot)$, we have a different context each time for independent measurements and instead we have

$f(x) = g_i(a)g_i(b)$
$f(y) = g_j(a)g_j(b')$
$f(z) = g_k(a')g_k(b)$
$f(w) = g_l(a')g_l(b')$

The factorizations do not work, so we are left with only the settings dependency present in the fourth term and must use the previous method to analyze the upper bound, which gives us

$3f(x) - f(3x)$

In summary:

- Counterfactual dependent terms like in Bell/CHSH gives us $g_i(a)[g_i(b) + g_i(b')] + g_i(a')[g_i(b) - g_i(b')] \leq 2$
- 4 Independent measurements with 3 independent angle variables gives us a maximum of $3f(x) - f(3x)$, which for f(x) = cos(x) is 2 sqrt 2. You can solve this equation to find what settings will give you the maximum values and it turns out to be just the ones used by Bell. But there are many other combinations. See the wolfram alpha link above.
- 4 Completely independent measurements at 4 independent settings, gives us a maximum of 4.
Therefore, "violation" is only possible, if the terms are independent but impossible for terms with the same dependence as in the CHSH.

Note that if the 4 variables $w, x, y, z$ are indepependent, even if $f(x) = cos(x)$, we will still have an upper bound of 4.
minkwe

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

minkwe wrote:Counterfactual dependent terms like in Bell/CHSH gives us $g_i(a)[g_i(b) + g_i(b')] + g_i(a')[g_i(b) - g_i(b')] \leq 2$

This is correct if $g_i(a)$ etc. are the commuting scalar numbers.

But if $g_i(a)$ are non-commuting numbers -- such as the standardized bivectors used in my derivation -- then the bound is $2\sqrt{2}$ , as verified by Fred using GAviewer.
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
minkwe wrote:Counterfactual dependent terms like in Bell/CHSH gives us $g_i(a)[g_i(b) + g_i(b')] + g_i(a')[g_i(b) - g_i(b')] \leq 2$

This is correct if $g_i(a)$ etc. are the commuting scalar numbers.

But if $g_i(a)$ are non-commuting numbers -- such as the standardized bivectors used in my derivation -- then the bound is $2\sqrt{2}$ , as verified by Fred using GAviewer.

If they are noncommuning then you have a situation more like

$f(x) = g_i(a)g_i(b)$
$f(y) = g_j(a)g_j(b')$
$f(z) = g_k(a')g_k(b)$
$f(w) = g_l(a')g_l(b')$

Which can not be factorized and left with only the angle dependences, you do indeed get 2 root 2. Non-commutation means the terms must be independent, since they can't be measured simultaneously.

It confirms the conclusion that "violation" is only possible with independent terms.
minkwe

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

minkwe wrote:If they are noncommuning then you have a situation more like

$f(x) = g_i(a)g_i(b)$
$f(y) = g_i(a)g_i(b')$
$f(z) = g_i(a')g_i(b)$
$f(w) = g_i(a')g_i(b')$

Which can not be factorized and left with only the angle dependences, you do indeed get 2 root 2. Non-commutation means the terms must be independent, since they can't be measured simultaneously.

It confirms the conclusion that "violation" is only possible with independent terms.

Didn't you mean,

$f(x) = g_i(a)g_i(b)$
$f(y) = g_j(a)g_j(b')$
$f(z) = g_k(a')g_k(b)$
$f(w) = g_l(a')g_l(b')$
?
Anyways, we know now that QM and Joy's derivation don't actually violate Bell-CHSH. However, I do think that Joy's derivation does match and explain the QM prediction physically.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

minkwe wrote:
Joy Christian wrote:
minkwe wrote:Counterfactual dependent terms like in Bell/CHSH gives us $g_i(a)[g_i(b) + g_i(b')] + g_i(a')[g_i(b) - g_i(b')] \leq 2$

This is correct if $g_i(a)$ etc. are the commuting scalar numbers.

But if $g_i(a)$ are non-commuting numbers -- such as the standardized bivectors used in my derivation -- then the bound is $2\sqrt{2}$ , as verified by Fred using GAviewer.

If they are noncommuning then you have a situation more like

$f(x) = g_i(a)g_i(b)$
$f(y) = g_i(a)g_i(b')$
$f(z) = g_i(a')g_i(b)$
$f(w) = g_i(a')g_i(b')$

Which can not be factorized and left with only the angle dependences, you do indeed get 2 root 2. Non-commutation means the terms must be independent, since they can't be measured simultaneously.

It confirms the conclusion that "violation" is only possible with independent terms.

This is correct too. However, I have always been able to have my cake and eat it too, which, sadly, not many people have understood . What do I mean by that?

Well, in my framework the $g_i(a)$ etc. in the above equations are standardized variables, or standard scores, not the experimentally observed raw scores. I have explained the important distinction in many papers [see, for example, Eqs. (105) to (111) of this paper] so I will not go through it again, but here is the relationship between the two [see Eqs. (7) and (8) of this paper]:

A_i(a, u) = d(a) h_i(a, u) = +/-1 (= commuting scalar number),

where u is the hidden variable, a is the measurement direction, d(a) is a detector bivector, h_i(a, u) is the non-commuting standard score (a bivector), and A_i(a, u) = +/-1 is the actually observed raw score. Since A_i(a, u) are commuting scalar numbers, when naively calculated [as Gill is shown to do in Eqs. (12) and (14) of this paper] the bound appears to be 2. However, when correctly calculated using the standardized variables h_i(a, u), which are non-commuting numbers, the bound is $2\sqrt{2}$. So here we have a situation where "totally dependent" raw scores A_i(a, u) (which are the same in both E11 and E12 of CHSH) leads to a bound greater than 2.

This of course matches perfectly with the standard experimental computation of the actual correlation, as shown in the derivation of Eq. (B10) in the above paper.

Unfortunately subtleties like these in my work has allowed people like Gill to twist facts and misrepresent my work, either because they are dimwits or dishonest.

PS: Here is a nice pedagogical explanation of the difference between raw scores and standard scores: http://www.gifted.uconn.edu/siegle/rese ... cores.html
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
minkwe wrote:If they are noncommuning then you have a situation more like

$f(x) = g_i(a)g_i(b)$
$f(y) = g_i(a)g_i(b')$
$f(z) = g_i(a')g_i(b)$
$f(w) = g_i(a')g_i(b')$

Which can not be factorized and left with only the angle dependences, you do indeed get 2 root 2. Non-commutation means the terms must be independent, since they can't be measured simultaneously.

It confirms the conclusion that "violation" is only possible with independent terms.

Didn't you mean,

$f(x) = g_i(a)g_i(b)$
$f(y) = g_j(a)g_j(b')$
$f(z) = g_k(a')g_k(b)$
$f(w) = g_l(a')g_l(b')$
?

Yes Fred that is exactly what I meant.
minkwe

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Joined: Sat Feb 08, 2014 10:22 am

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

So now that we are done with Joy's derivation, let's take a look at the QM derivation.

https://en.wikipedia.org/wiki/Quantum_m ... prediction

I think from that we can get the following "CHSH" type expression for 1/2 spin quantum objects.

$\cos(a-b)+\cos(a-b')+\cos(a'-b)-\cos(a'-b')\leq 2 \sqrt{2}$

Then we can show that (a-b')+(a'-b)-(a-b) = a'-b' so that we have the situation that Michel demonstrated.

And actually Joy's model could be done this way also since his model predicts E(a, b) = -a.b.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:So now that we are done with Joy's derivation, let's take a look at the QM derivation.

https://en.wikipedia.org/wiki/Quantum_m ... prediction

I think from that we can get the following "CHSH" type expression for 1/2 spin quantum objects.

$\cos(a-b)+\cos(a-b')+\cos(a'-b)-\cos(a'-b')\leq 2 \sqrt{2}$

Then we can show that (a-b')+(a'-b)-(a-b) = a'-b' so that we have the situation that Michel demonstrated.

And actually Joy's model could be done this way also since his model predicts E(a, b) = -a.b.

PS: Also, my original 2007 derivation of the $2\sqrt{2}$ bound is similar to the one in this Wikipedia article. I had all of the key derivations already done in 2007.
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:So now that we are done with Joy's derivation, let's take a look at the QM derivation.

https://en.wikipedia.org/wiki/Quantum_m ... prediction

I think from that we can get the following "CHSH" type expression for 1/2 spin quantum objects.

$\cos(a-b)+\cos(a-b')+\cos(a'-b)-\cos(a'-b')\leq 2 \sqrt{2}$

Then we can show that (a-b')+(a'-b)-(a-b) = a'-b' so that we have the situation that Michel demonstrated.

And actually Joy's model could be done this way also since his model predicts E(a, b) = -a.b.

PS: Also, my original 2007 derivation of the $2\sqrt{2}$ bound is similar to the one in this Wikipedia article. I had all of the key derivations already done in 2007.

Yes. But the point I was making is how Michel's demonstration applies to the QM derivation and to yours.

$\cos(x)+\cos(y)+\cos(z)-\cos(y+z-x)\leq 2 \sqrt{2}$

IOW, the dependency in the (a'-b') = y+z-x
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

Man, these Codecogs tex graphics are really coming in slow for me.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:Fred, Please see Eq. (5) of my Reply to Critics.

PS: Also, my original 2007 derivation of the $2\sqrt{2}$ bound is similar to the one in this Wikipedia article. I had all of the key derivations already done in 2007.

Yes. But the point I was making is how Michel's demonstration applies to the QM derivation and to yours.

$\cos(x)+\cos(y)+\cos(z)-\cos(y+z-x)\leq 2 \sqrt{2}$

IOW, the dependency in the (a'-b') = y+z-x

Fair enough. I think Michel's argument does go through, at least for the special (non 3D vector) case of (a - b) etc., and that is sufficient for his argument.
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:
Joy Christian wrote:Fred, Please see Eq. (5) of my Reply to Critics.

PS: Also, my original 2007 derivation of the $2\sqrt{2}$ bound is similar to the one in this Wikipedia article. I had all of the key derivations already done in 2007.

Yes. But the point I was making is how Michel's demonstration applies to the QM derivation and to yours.

$\cos(x)+\cos(y)+\cos(z)-\cos(y+z-x)\leq 2 \sqrt{2}$

IOW, the dependency in the (a'-b') = y+z-x

Fair enough. I think Michel's argument does go through, at least for the special (non 3D vector) case of (a - b) etc., and that is sufficient for his argument.

Yeah, mostly QM and the experiments are only concerned with 2D a and b anyways. But we saw with your long derivation that it works fully 3D and GAViewer confirms it.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:But we saw with your long derivation that it works fully 3D and GAViewer confirms it.

Actually, it would be nice to make a single post here showing the full derivation step by step, from the beginning to end, using the GAviewer code. You already have various steps and the code in your earlier posts in this thread, but that may be confusing for readers. A single post could also help in citing the result in the future.
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:But we saw with your long derivation that it works fully 3D and GAViewer confirms it.

Actually, it would be nice to make a single post here showing the full derivation step by step, from the beginning to end, using the GAviewer code. You already have various steps and the code in your earlier posts in this thread, but that may be confusing for readers. A single post could also help in citing the result in the future.

Yeah, I think a main part that is missing is showing how 4(a1 x a2).(b2 x b1) matches the scalar part of,

- A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 ,

using GAViewer. I only stated it and did not show it. Here it is.

Code: Select all
>> -(A2 A1 B2 B1) + (A2 A1 B1 B2) + (A1 A2 B2 B1) - (A1 A2 B1 B2)ans = 0.47 + -0.62*e2^e3 + -0.01*e3^e1 + 0.22*e1^e2>> t=crosspr(a1, a2)t = -0.07*e1 + 0.14*e2 + -0.19*e3>> q=crosspr(b2, b1)q = 0.25*e1 + 0.27*e2 + 0.73*e3>> -4*t.qans = 0.47

So we can see that the scalars, 0.47, from both procedures match. Here is the complete GAViewer code that sets up doing the manual entries and shows the cross product function. This is adapted from Albert Jan Wonnink's original code.
Code: Select all
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;           a1=getRandomUnitVector();     b1=getRandomUnitVector();     a2=getRandomUnitVector();     b2=getRandomUnitVector();      A1=-I.a1;      A2=-I.a2;      B1=I.b1;      B2=I.b2;      C=((A1 B1)+(A1 B2)+(A2 B1)-(A2 B2));      D=C*C;      print(C);      print(D);}

So I will try to do a complete compilation tomorrow. Unfortunately, GAViewer won't do the comutation of the A's and B's part but that is easy to show by hand as already done.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

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

Fred,

If you use GAViewer to calculate (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 with your trial bivectors, do you get the same answer for both? If not, what is your explanation for the difference?

Do you remember your "triumph" of demonstrating left handed bivectors in GAViewer by commuting product order of the bivectors? The result was a negation of the bivector component of AB, was it not? What does this tell you about [A, B]? Would this not be twice the bivector component of AB? With unit bivectors A and B, would [A, B} then = 0 if and only if A = B?
Rick Lockyer

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

Rick Lockyer wrote:Fred,

If you use GAViewer to calculate (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 with your trial bivectors, do you get the same answer for both? If not, what is your explanation for the difference?

Do you remember your "triumph" of demonstrating left handed bivectors in GAViewer by commuting product order of the bivectors? The result was a negation of the bivector component of AB, was it not? What does this tell you about [A, B]? Would this not be twice the bivector component of AB? With unit bivectors A and B, would [A, B} then = 0 if and only if A = B?

As far as I currently know, GAViewer can't do (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) properly to get 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 because it doesn't know about Ax's commuting with Bx's where x = 1 or 2. So that part has to be done by hand as shown above. Someone more clever than I might know a way of doing it though.

It wasn't my "triumph". It was Albert Jan Wonnink's. The answer to your last two questions is no. Look carefully at the terms. We have here A1, A2, B1, B2.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

Rick Lockyer wrote:Fred,

If you use GAViewer to calculate (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 with your trial bivectors, do you get the same answer for both? If not, what is your explanation for the difference?

Do you remember your "triumph" of demonstrating left handed bivectors in GAViewer by commuting product order of the bivectors? The result was a negation of the bivector component of AB, was it not? What does this tell you about [A, B]? Would this not be twice the bivector component of AB? With unit bivectors A and B, would [A, B} then = 0 if and only if A = B?

For those readers who are unaware, all of the misguided arguments of Rick Lockyer have been thoroughly refuted here: viewtopic.php?f=6&t=183&start=110#p5107.

He has, however, learned nothing from his humiliation. He continues his ignorant arguments without understanding the first thing about the physics involved in Fred's demonstration above. His mistake again is to not realize that (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 cannot give the same answer because the first factor does not respect the physical fact that A's and B's commute because they are space-like separated. This is the so-called mathematician's folly. My advice to him is to stop humiliating himself further and learn some basic physics before trying to understand my work.

For other readers, I would urge to actually read my two papers referred to by Fred in his post above.
Joy Christian
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Rick Lockyer wrote:Fred,

If you use GAViewer to calculate (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) and 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 with your trial bivectors, do you get the same answer for both? If not, what is your explanation for the difference?

Do you remember your "triumph" of demonstrating left handed bivectors in GAViewer by commuting product order of the bivectors? The result was a negation of the bivector component of AB, was it not? What does this tell you about [A, B]? Would this not be twice the bivector component of AB? With unit bivectors A and B, would [A, B} then = 0 if and only if A = B?

As far as I currently know, GAViewer can't do (A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) properly to get 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 because it doesn't know about Ax's commuting with Bx's where x = 1 or 2. So that part has to be done by hand as shown above. Someone more clever than I might know a way of doing it though.

It wasn't my "triumph". It was Albert Jan Wonnink's. The answer to your last two questions is no. Look carefully at the terms. We have here A1, A2, B1, B2.

And if you look previously in this thread you will see that Joy's local realistic model can do the derivation in the same way that QM does since E(a, b) = -a.b in his model. So there is no doubt that the result of $2 \sqrt{2}$ is correct. This longer derivation is a good example of geometric algebra in action for the model.

And... I forgot to mention in the long derivation that the vectors, a1, a2, b1, and b2 are all fully 3D in the GAViewer validation as opposed to 2D vectors that are used in the QM derivation.
FrediFizzx
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### Re: Joy's derivation of the 2*sqrt(2) bound

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.

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.

Joy, "misguided", "humiliation"? Really? I apologize for saying I have lost respect for you. After these responses, I just feel sorry for you. This is extremely simple math, and you clearly have it wrong. After the educational accomplishments you have had, if you cannot see this, I worry there may be something seriously wrong with you.
Rick Lockyer

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