Are Gill and Moldoveanu disingenuous or incompetent?

[Joy Christian]

It is not my place to evaluate the merit of Joy Christian’s arguments against Bell’s theorem, but what is clear is the enormous amount of time and energy wasted arguing, debating, attacking and defending positions. How many years has it been?

The way I see it, if Joy Christian is satisfied with his disproof of Bell’s theorem and confident about it then he should work on something else rather than wasting even a minute in petty and useless debates. If right, his disproof will at some point it gather acceptance. The question will be settled one way or another as it always is and must be; through observations and experiments.
Even if no one actually performs the experiment proposed by Christian, other experiments will indirectly falsify or support it eventually.

The point is, Joy Christian job’s in regards to Bell’s theorem is done (unless he decides to perform the experiment himself). The time wasted on war waged to prove something that cannot be proved without experiments would have been better used working on a new problem, on doing actual physics. Anything else is in my opinion a terrible waste of talent.

Daniel,

The important issue for me is not so much about Bell's theorem -- which was stillborn to begin with -- but about understanding the origins of the quantum correlations. I have proved a very general theorem, which shows how ALL quantum correlations (i.e., not just the EPR-Bohm correlation) can be understood local-realistically. This theorem does point to exciting new physics, and we are working on that. But we are not ready yet to talk about that in public. The petty war of words you refer to is just a sideshow, an amusement if you like. I don't take the critics or their arguments seriously. As far as I am concerned they don't know what they are talking about.

[FrediFizzx]

It is not my place to evaluate the merit of Joy Christian’s arguments against Bell’s theorem, but what is clear is the enormous amount of time and energy wasted arguing, debating, attacking and defending positions. How many years has it been?

If you are not interested in evaluating Joy's arguments, then why are you even commenting about this? What would you do if there was a concerted effort to lie about and misrepresent your work constantly?

The way I see it, if Joy Christian is satisfied with his disproof of Bell’s theorem and confident about it then he should work on something else rather than wasting even a minute in petty and useless debates. If right, his disproof will at some point it gather acceptance. The question will be settled one way or another as it always is and must be; through observations and experiments.
Even if no one actually performs the experiment proposed by Christian, other experiments will indirectly falsify or support it eventually.

Microscopically, there is no experiment that can falsify Joy's local realistic model without falsifying quantum mechanics (it's not going to happen, IMHO). Something the critics completely miss. The quantum experiments actually support the model microscopically. However, if a proper macroscopic singlet type experiment were successful, then that would put this to bed once and for all. The critics are doing everything they can to try to suppress this experiment from being done by lying and total misrepresentation constantly.

The point is, Joy Christian job’s in regards to Bell’s theorem is done (unless he decides to perform the experiment himself). The time wasted on war waged to prove something that cannot be proved without experiments would have been better used working on a new problem, on doing actual physics. Anything else is in my opinion a terrible waste of talent.

Sometimes these debates and arguments lead to refinements. So not a waste of time.

[Joy Christian]

Richard Gill has struck again. He has now conspired to have Paul Snively's Blog closed down to save himself exposed as an extreme Bell fanatic and a compulsive liar.

But fortunately I have been able to retrieve the important last post by Snively from a google catch of his former blog. Let me reproduce it here for the record:

Note to everyone still reading,

I'm closing comments on this thread.

Let me reiterate that it's been a very rewarding, educational discussion, even in instances when there has been vociferous disagreement among its participants. It has underscored my appreciation of constructivist mathematics, quantum mechanics, and geometric algebra in particular.

However, I have had to ban Dr. Gill from my blog, and even so his proxies have taken it upon themselves to continue to post here. In addition, Dr. Gill is now e-mailing me directly, when I've made quite clear that my interest in Bell's theorem is strictly tangential. Further, he continues to reject the outcome of the computational process that has demonstrated the success of Dr. Christian's model unambiguously.

From here there is literally nowhere to go. I cannot help anyone who asserts the primacy of their opinion over a computed demonstration. The comments, the links, all of it will remain for as long as Disqus is generous enough to host it. But this is now the end of the line for this thread.

Thanks to all of the participants, including Dr. Gill. But please stop e-mailing me.

Of course, Gill might try next to close down this forum as well, but we shall see whether he succeeds. [Paul Snively is referring to this computed demonstration.]

[Ben6993]

Joy Christian wrote on Sun Jul 26, 2015 2:45 pm

Ben6993 wrote:
My wife received last week a metal bracelet in the form of a Moibus strip birthday present.

That is the best piece of news / information, Ben, in all of these threads put together. Belated Happy Birthday to your wife. I was also born in July, by the way.

Hi Joy, Thanks for the kind thoughts, My wife and I are both in Leo but in different months.

By a coincidence, an issue of a reversed sign is not new to me. My, now late and lamented, ex-boss used some of my raw data in his Ph.D thesis twenty or so years ago. In advance, he showed an acknowlegement to me in the appropriate place in the thesis but I was only allowed to read the thesis after publication. Then I found a chunk of his analysis where he had made an sign error. It was a gender (B and G) by performance (H and L) statistic and he had calculated (AH-BH)-(AL-BL) but got the sign wrong which meant that some of his conclusions were pointing in the wrong direction! However, he corrected it and even went on to get another paper out of it. Identifying that error was quite straightforward.

I can't help thinking that the sign issue in the current case is now getting close to resolution.

Best wishes

[Joy Christian]

I can't help thinking that the sign issue in the current case is now getting close to resolution.

There never was a sign issue, Ben. None whatsoever. And we now even have a computed demonstration of the correctness of my model, as Paul Snively rightly put it.

[minkwe]

Richard Gill has struck again. He has now conspired to have Paul Snively's Blog closed down to save himself exposed as an extreme Bell fanatic and a compulsive liar.

https://twitter.com/nastassiadavis/stat ... 1254482944
https://disqus.com/home/discussion/psni ... s_blog_53/

[Rick Lockyer]

I can't help thinking that the sign issue in the current case is now getting close to resolution.

Best wishes

Yes Ben, we all hope it will be resolved soon. For those not believing in an isomorphism between algebras, here is a snippet of the product in right and left GA straight up:

Right handed:
(a1 e2^e3 + a2 e3^e1) ^ (b1 e2^e3 + b2 e3^e1)
=
(a1 b1 e2^e3^e2^e3 + a2 b2 e3^e1^e3^e1)
+ (a1 b2 e2^e3^e3^e1 + a2 b1 e3^e1^e2^e3)
=
(– a1 b1 – a2 b2)
+ (a1 b2 – a2 b1) e1^e2

Left handed:
(a1 e3^e2 + a2 e1^e3) ^ (b1 e3^e2 + b2 e1^e3)
=
(a1 b1 e3^e2^e3^e2 + a2 b2 e1^e3^e1^e3)
+ (a1 b2 e3^e2^e1^e3 + a2 b1 e1^e3^e3^e2)
=
(– a1 b1 – a2 b2)
– (a1 b2 – a2 b1) e2^e1 sign reversal on coefficient BUT left handed basis still
=
(– a1 b1 – a2 b2)
+ (a1 b2 – a2 b1) e1^e2 an equivalence with right handed

I would hope this is simple enough for ALL of Joy’s supporters to understand. There is no sign change on the cross product when placed in an additive form (same basis element to permit sum of coefficients)

Sadly, full resolution will only come about when Joy and his followers man up and admit their mistakes, and apologize to those they have slighted. Not holding my breath.

[Joy Christian]

Richard Gill has struck again. He has now conspired to have Paul Snively's Blog closed down to save himself exposed as an extreme Bell fanatic and a compulsive liar.

https://twitter.com/nastassiadavis/stat ... 1254482944
https://disqus.com/home/discussion/psni ... s_blog_53/

Ah... Good to know that Gill wasn't able to have the Disqus part of Paul's blog closed down. Thanks, Michel.

As for the junk Lockyer keeps posting, we should have a rubbish bin on this forum where we can dump such garbage.

[minkwe]

I would hope this is simple enough for ALL of Joy’s supporters to understand. There is no sign change on the cross product when placed in an additive form (same basis element to permit sum of coefficients)
Sadly, full resolution will only come about when Joy and his followers man up and admit their mistakes, and apologize to those they have slighted. Not holding my breath.

Rick, a computation has been provided which produces what you say is impossible. Perhaps you will be convinced if you understood how and why Bell's theorem is deeply flawed, as I have shown here: viewtopic.php?f=6&t=181 , here: viewtopic.php?f=6&t=168&start=230#p4731, and and here viewtopic.php?f=6&t=75&start=170#p4380

So if you are in doubt, hop on over to viewtopic.php?f=6&t=181, and answer the question I've been asking Bell's followers and getting a blank stare:

Here is the question: Please demonstrate that QM violates Bell's inequality

by providing , the QM predictions for the terms

Once it becomes clear that Bell's theorem is false, it will also become evident that QM is not incompatible with Local realistic models and that opens up the mind to consider possibilities like Joy's model. Note that there is often a subtle error in response to that question. Errors also often made by Joy's detractors, which Fred aptly characterized as follows:

[they], are hung up on some *algebraic* outcomes of a *statistical* model.

But I'll wait for your answers in the other thread.

[FrediFizzx]

Richard Gill has struck again. He has now conspired to have Paul Snively's Blog closed down to save himself exposed as an extreme Bell fanatic and a compulsive liar.

https://twitter.com/nastassiadavis/stat ... 1254482944
https://disqus.com/home/discussion/psni ... s_blog_53/

Ah... Good to know that Gill wasn't able to have the Disqus part of Paul's blog closed down. Thanks, Michel.

As for the junk Lockyer keeps posting, we should have a rubbish bin on this forum where we can dump such garbage.

Here is the first part of Paul's blog from Jan., 2015

http://web.archive.org/web/201501230153 ... 2/Fallacy/

[Ben6993]

Hi Rick

Relying in haste before going out and while the thread is still available.

I know very little GA but I am not sure if anyone is going to respond from joy's side and I would like to keep the discussion going.

I have followed all your last post up to the last two line of formulae. Just as a maths exercise not related to the GA program.

AFAIK, and that isn't very much, those last two line follow on as an exact equality within the LH basis. I am not sure why you are mentioning the RH equivalence?

e2^e1 =-e1^e2 within the LH basis doesn't it?

To convert the bases to a different e1e2e3 handedness would need more than that wouldn't it?

[FrediFizzx]

Hi Rick

Relying in haste before going out and while the thread is still available.

I know very little GA but I am not sure if anyone is going to respond from joy's side and I would like to keep the discussion going.

I have followed all your last post up to the last two line of formulae. Just as a maths exercise not related to the GA program.

AFAIK, and that isn't very much, those last two line follow on as an exact equality within the LH basis. I am not sure why you are mentioning the RH equivalence?

e2^e1 =-e1^e2 within the LH basis doesn't it?

To convert the bases to a different e1e2e3 handedness would need more than that wouldn't it?

Ah... this kind of reminds me as to what Albert Jan, Joy and I were going through when Albert Jan came up with the correct solution.

http://challengingbell.blogspot.com/201 ... of_30.html

[Rick Lockyer]

Hi Rick

Relying in haste before going out and while the thread is still available.

I know very little GA but I am not sure if anyone is going to respond from joy's side and I would like to keep the discussion going.

I have followed all your last post up to the last two line of formulae. Just as a maths exercise not related to the GA program.

AFAIK, and that isn't very much, those last two line follow on as an exact equality within the LH basis. I am not sure why you are mentioning the RH equivalence?

e2^e1 =-e1^e2 within the LH basis doesn't it?

To convert the bases to a different e1e2e3 handedness would need more than that wouldn't it?

Ben,

To be able to relate the left and right handed GA products, we must stick with identical scalar and vector basis elements. So for both we must have {1, e1, e2, e3}.

The left handed bivector basis elements are built from the product of the orientation determining trivector basis element +I and {e1, e2, e3} where I = e1^e2^e3.

We have for the three left handed bivector basis elements

I^e1 = e1^I = - e2^e3 = e3^e2 call this L1
I^e2 = e2^I = - e3^e1 = e1^e3 call this L2
I^e3 = e3^I = - e1^e2 = e2^e1 call this L3

The right handed bivector basis elements are built from the product of the orientation determining trivector basis element - I and {e1, e2, e3} where

- I = - e1^e2^e3 = e1^e3^e2 = e2^e1^e3 = e3^e2^e1

We have for the three right handed bivector basis elements

- I^e1 = e1^- I = e2^e3 = - e3^e2 call this R1
- I^e2 = e2^- I = e3^e1 = - e1^e3 call this R2
- I^e3 = e3^- I = e1^e2 = - e2^e1 call this R3

Of course the right and left names, and the definition of I are arbitrary. The above choices allow e1, e2 and e3 to map to familiar x, y and z physical coordinates, allowing the right handed bivector bases to map to the right hand rule for vector cross products in e.

To answer one of your questions, the wedge product always anti-commutes.

We have clearly R1 = - L1, R2 = - L2 and R3 = - L3. Also

L1^L2 = - L3 = - L2^L1
L2^L3 = - L1 = - L3^L2
L3^L1 = - L2 = - L1^L3

R1^R2 = R3 = - R2^R1
R2^R3 = R1 = - R3^R2
R3^R1 = R2 = - R1^R3

If you look up the multiplication rules for quaternions, you will find the above multiplication rules are identical to those for quaternions if we assign quaternion basis sets {1, R1, R2, R3} and {1, L1, L2, L3} of course now not really wedge products, just wedge equivalent results. This is why quaternion algebra is isomorphic to the bivector – scalar basis algebra. I used quaternions in my proof above because it involves less typing.

So what is the bottom line here? The sign difference in the Lx^Ly = - Lz verses Rx^Ry = +Rz gives one sign change to all of the coefficients with bivector result bases in the same handedness. The sign change Joy missed is the one required to map between Left and Right in order to have the same basis element handedness on all terms, the ONLY way they can be combined (added here). So two negations is no negation, and we have an equivalence. Another way of looking at it is he is only correct if +1 = -1. The complaints by me and others are quite legitimate.

You are correct, Joy will probably say nothing, or say that I do not know what I am talking about. One thing we will never see out of him is a baby-step proof of his claim there is an orientation change conjugation that allows the non-scalar sums to tend to zero in the limit with his program. Nor will you get a explicit statement of WHERE any of my above analysis is not correct, just blanket statements.

Hope this is clear enough for you.

[Rick Lockyer]

I would hope this is simple enough for ALL of Joy’s supporters to understand. There is no sign change on the cross product when placed in an additive form (same basis element to permit sum of coefficients)
Sadly, full resolution will only come about when Joy and his followers man up and admit their mistakes, and apologize to those they have slighted. Not holding my breath.

Rick, a computation has been provided which produces what you say is impossible. Perhaps you will be convinced if you understood how and why Bell's theorem is deeply flawed, as I have shown here: viewtopic.php?f=6&t=181 , here: viewtopic.php?f=6&t=168&start=230#p4731, and and here viewtopic.php?f=6&t=75&start=170#p4380

So if you are in doubt, hop on over to viewtopic.php?f=6&t=181, and answer the question I've been asking Bell's followers and getting a blank stare:

Here is the question: Please demonstrate that QM violates Bell's inequality

by providing , the QM predictions for the terms

Once it becomes clear that Bell's theorem is false, it will also become evident that QM is not incompatible with Local realistic models and that opens up the mind to consider possibilities like Joy's model. Note that there is often a subtle error in response to that question. Errors also often made by Joy's detractors, which Fred aptly characterized as follows:

[they], are hung up on some *algebraic* outcomes of a *statistical* model.

But I'll wait for your answers in the other thread.

I am not sure why you conflate people that do not agree with Joy Christian with supporters of Bell. Did you ever consider that maybe both are wrong, or Joy's claims have no bearing on Bell for all of its intentions? For the record, I could not care less about Bell's theorem, it does not adequately address the physics. For that matter neither does Joy's work. Where is the model for the particles, for the detectors and how they do their job? Both pieces are math exercises with vague connections to physics, little more. I would put Joy's more vague than Bell.

Funny how this thread was started by Joy to dump on detractors who claim his bivector orientation fair coin analysis is f'd up, yet when I demonstrate it REALLY is we get idiotic statements like the above from Fred about being hung up on the algebra, which is PRECISELY what this thread is all about. Wow! So where do you come down on this? Joy have it right? Then you need to prove me wrong or say you really do not know if he is right or wrong. Good luck with that.

[minkwe]

I thought I made it clear in my previous posts that the computation disputes what you are saying. The bi-vector part does vanish contrary to your claims, as the posted code shows. You will have to find some other criticism because this one is not convincing at all.

The sign change Joy missed is the one required to map between Left and Right in order to have the same basis element handedness on all terms, the ONLY way they can be combined (added here). So two negations is no negation, and we have an equivalence. Another way of looking at it is he is only correct if +1 = -1. The complaints by me and others are quite legitimate.

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.

You say you aren't interested in Bell's theorem but I asked you to consider looking more deeply into it because a similar confusion between algebraic and statistical computations lie at the core of it. You aren't interested, too bad. Fred's statement is spot on, and it is your characterization of it that looks completely misguided to me. So if you want to understand how correct algebra can be combined with woeful statistics and used to produce garbage, come over to the other thread and consider answering the question I asked.

[Rick Lockyer]

Rick,
I thought I made it clear in my previous posts that the computation disputes what you are saying. The bi-vector part does vanish contrary to your claims, as the posted code shows. I don't see where you are going with this.

You say you aren't interested in Bell's theorem but I do not conflate anything. I asked you to consider looking more deeply into it because a similar confusion between algebraic and statistical computations lie at the core of it. You aren't interested, too bad. Fred's statement is spot on, and it is your characterization of it that looks completely misguided to me. So if you want to understand how correct algebra can be combined with woeful statistics and used to produce garbage, come over to the other thread and consider answering the question I asked.

Your issues with Bell are disjoint to the question about the bivector part vanishing. Could not find any justification to your claims it does in any of the topics you linked. All code I have looked at takes Joy's mistake as a given, so their results are meaningless. This is strictly an algebraic question, with absolutely no dependencies on whether or not Bell is correct, whether or not S^3 applies to Bell and non-locality. So stop the obfuscation. Since it is purely an algebraic mistake by Joy, and I have given you the proof, show me where I made a mistake. Put up or shut up.

[FrediFizzx]

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?

Alice and Bob are always in the same basis per particle pair (iteration or trial). The way the GAViewer code works (and I am sure that you already know this so presenting it for interested lurkers) is that if lambda = +1 then their geometric product AB is in the right handed basis. If lambda = -1, then their geometric product AB is in the left handed basis. Now, what does a left handed system look like from a right handed only perspective? The geometric product order is reversed to BA. Now Lockyer is claiming there is some kind of mistake even before this. But how can there be as all Joy's equations work perfectly in GAViewer and Python and give the right result? He can't even really point to the exact place in Joy's math where he thinks it is. Here are the equations in the GAViewer code.

C=-I.a; //C = {-a_j B_j}
D=I.b; //D = {b_k B_k}
E=mu.a; //E = {a_k B_k(L)}
F=mu.b; //F = {b_j B_j(L)}
A=C E; //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
B=F D; //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}

Where exactly could there be a mistake? There is none. It's way too simple for there to be a mistake.

[Joy Christian]

I find it amusing that our schoolboy simply ignored the crucial point made by Michel; for if he faced up to it then he would be running home with tail between his legs:

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 for the proof of my argument, there are fifteen papers, a book, and literally thousands of my posts on the internet at ones disposal, in particular these two papers:

http://arxiv.org/abs/1211.0784

and

http://arxiv.org/abs/1501.03393.

[Rick Lockyer]

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?

Alice and Bob are always in the same basis per particle pair (iteration or trial). The way the GAViewer code works (and I am sure that you already know this so presenting it for interested lurkers) is that if lambda = +1 then their geometric product AB is in the right handed basis. If lambda = -1, then their geometric product AB is in the left handed basis. Now, what does a left handed system look like from a right handed only perspective? The geometric product order is reversed to BA. Now Lockyer is claiming there is some kind of mistake even before this. But how can there be as all Joy's equations work perfectly in GAViewer and Python and give the right result? He can't even really point to the exact place in Joy's math where he thinks it is. Here are the equations in the GAViewer code.

C=-I.a; //C = {-a_j B_j}
D=I.b; //D = {b_k B_k}
E=mu.a; //E = {a_k B_k(L)}
F=mu.b; //F = {b_j B_j(L)}
A=C E; //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
B=F D; //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}

Where exactly could there be a mistake? There is none. It's way too simple for there to be a mistake.

One more time, for those like Fred who are having an inconceivable amount of trouble (more like reluctance and/or denial) getting it. This is extremely elementary algebra.

The coefficients for the non-scalar terms here absolutely do change signs just as you claim when the product is made in the other handedness system. I have never had a problem with this, and that CLEARLY is indicated in my proof above, my response to Ben, everywhere. The rub is, these coefficients are attached to basis elements with opposite handedness, so the coefficients cannot be directly added, but Joy and you by blindly following him do just that. Before adding, either right non-scalar bases need to be mapped to left, or left to right. then and ONLY then may the coefficients be combined. Either of these mappings does an ADDITIONAL NEGATION YOU ALL HAVE NEGLECTED. When put into an addable form, the coefficients have the same sign.

Once again Fred, as I have repeatably told you, reversing order changes the sign of these coefficients, true enough. You have assumed this is what you needed to do prior to adding because you have bought into Joy's position the signs change and can then be added. In all of your code, you assumed what you say you proved. Since what you assumed was incorrect, you have proved nothing.

[Joy Christian]

Since our schoolboy is determined to endlessly spam this thread with his absolute and utter garbage that has nothing to do with either my model or the physics and mathematics of the EPR-Bohm correlation, I too will indulge in spamming the thread and reproduce what I have already said before in several different ways by now:

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.