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FQXi BLOGS
April 23, 2018

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TOPIC: Part 3 of To Be on Not to Be (a Local Realist) [refresh]
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Blogger Florin Moldoveanu wrote on Sep. 5, 2011 @ 10:19 GMT
Hey everyone,

I really enjoyed debating here Joy Christian’s challenge to Bell’s theorem, and this led me to take a more in-depth look at his results in an attempt to construct a computer simulation for his model. The simulation started with the Java script simulation of Cristi Stoica posted here earlier at FQXi and was augmented by free play experimentation into several models exhibiting the minus cosine quantum mechanics correlations. Those models gave me and Cristi confidence that we are on the right track and we should succeed into realizing numerically the same minus cosine correlations for Joy’s model as well. To do that I had to systematically double check the mathematical results in order to model it correctly as computer code. To my greatest surprise I discovered several mathematical errors which invalidates Joy Christian’s model and more importantly all its potential Clifford algebra generalizations. I have detailed the errors extensively in a separate paper which I posted to the arXiv:quant-ph/1109.0535 and I invite you to look at them there, since I do not want to write a long post with detailed maths here. I think these errors are serious enough to make the earlier debate that we have been having about the implications of the findings somewhat moot. I want to thank Cristi Stoica for all his help during the entire process of writing the paper.

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Joy Christian wrote on Sep. 5, 2011 @ 10:44 GMT
Dear Florin,

With sadness, I see that you have continued to ignore my latest papers and depended on the older, outdated debates. I will of course write a detailed rebuttle, but let me just note here that I do not agree with the so-called mistakes in my paper you mention.

Best,

Joy

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Florin Moldoveanu wrote on Sep. 5, 2011 @ 10:59 GMT
Dear Joy,

I am looking forward to your rebuttal. Please do read in detail my archive paper and see that unfortunately the mistakes are genuine and unrecoverable from. I have more than triple checked them and I would not write this paper on something this important if I were not completely sure. If I were to pick a single error, the first one would be it because during the change of handedness the Levi-Civita pseudo tensor should change signs as well (the cross product of two vectors is a pseudo-vector and not a vector) and this makes all the difference in eliminating the second term in the correlation.

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Joy Christian wrote on Sep. 5, 2011 @ 11:27 GMT
Dear Florin

I know the sincerity with which you have analyzed my work, and I do appreciate your interest in it. Unfortunately, you have again depended on my earlier papers, written four years ago, which are less than perfect, but not wrong. The trouble is that, as I have said before, you have not understood the difference between the raw scores and standard scores, and the corresponding difference between the actually observed values and the mathematical tools for evaluating the correlations between the actually observed values. Find me a convincing error in my one-page paper and I will give up my program. That one-page paper by itself is sufficient to vindicate EPR, disprove Bell, and show that quantum mechanics is an incomplete theory of nature. The rest are details, as Einstein would have put it.

Joy

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doug replied on Feb. 20, 2013 @ 01:32 GMT
The debate on this is way over my head, but, I suggest everyone bring in more experts. However, as regards,

"That one-page paper by itself is sufficient to vindicate EPR, disprove Bell, and show that quantum mechanics is an incomplete theory of nature."

Note: Of course quantum theory is incomplete. It needs CIG Theory !

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Florin Moldoveanu wrote on Sep. 5, 2011 @ 11:55 GMT
Dear Joy,

Sure, here it is. In http://arxiv.org/PS_cache/arxiv/pdf/1103/1103.1879v1.pdf please take a look at Eq. 4. It is incorrect. It has an illegal additional lambda term in RHS which does not belong there. Please explain how you arrive at Eq.~4 starting from Eq.~3. I guess you just do a substitution for beta_i to beta_i (lambda) and simplify lambda square of LHS. This is wrong however. You need an additional lambda for epsilon_ijk which would cancel the illegal lambda as well. Why is that? In any mathematical identity like Eq.3 you cannot simply swap mathematical terms. Your Eq. 4 is derived from Eq.3 by an actual symmetry which in general does preserve its mathematical validity. Specifically, your symmetry here is a mirror reflection which changes handedness. As such you need to apply this symmetry to ALL terms in Eq. 3 and this introduces an additional lambda for the Levi-Civita PSEUDO-tensor. Pseudo means an additional lambda term there. Derived correctly, Eq. 4 is identical with Eq. 3.

There is an alternative way to see the error. Start with LHS of Eq. 4, replace beta_j(lambda) beta_k(lambda) with beta_j beta_k lambda lambda = beta_j beta_k and then use Eq. 3 to replace beta_j beta_k with RHS of Eq. 3. Then compare this with RHS of Eq. 4. They are not equal (which they should be).

Florin

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Joy Christian replied on Sep. 5, 2011 @ 13:16 GMT
Dear Florin,

I posted this on a wrong thread before:

My equation (4) is not incorrect. It is a postulate, not a deduction from equation (3). A postulate cannot be incorrect. The motivation behind it comes from the already existing sign ambiguity discussed by Eberlein for example (cited in some of my papers). I strongly urge you to study this paper to recognize where you have gone wrong. You have got your mathematics all wrong, because you have not understood what the hidden variable in my model is. It is the handedness of the basis bivectors, or the orientation of the 3-sphere. This orientation "fluctuates" from right to left and back.

There is an easy way to bring out the running error in your paper. Whatever notations and conventions are used, the following two constraints on the basis bivectors must be satisfied:

For the right-handed basis bivectors one must maintain

(B_i)(B_j)(B_k) = +1 .

And for the left-handed basis bivectors one must maintain

(B_i)(B_j)(B_k) = -1 .

The hidden variable is then these two alternative possibilities.

If you maintain these constrains as consistency conditions throughout, then you will not be confused as you have been. And you certainly have been. Please read the Eberlein paper thoroughly.

Now, as you can easily check, the above two constraints are rigorously maintained in all of my papers, whatever the notations used.

On the other hand, these constraints are not satisfied by some of the crucial equations in your paper. Thus, I am afraid, your whole analysis is fundamentally flawed. By posting your fallacious analysis on the arXiv you have made my already difficult task much more difficult. Sadly, you are not helping physics by doing this. You are killing it off prematurely.

Best,

Joy

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Florin Moldoveanu replied on Sep. 5, 2011 @ 13:36 GMT
Dear Joy,

Let me copy my reply on this thread as well.

You can certainly have the freedom to choose any postulates you want, provided you do not choose inconsistent ones (because you can then derive any statement and its negation at the same time). As a postulate, Eq. 4 is inconsistent with Eq. 3 (Eq. 3 it is true). LHS of Eq. 3 is equal with LHS of Eq. 4 and therefore their right hand sides should be equal as well. Equating the RHS of Eq. 3 and 4 and simplifying the Kroneker delta and the Levi Civita results in: lambda = 1 in contradiction with your lambda = plus and minus one. In fact the inconsistency reduces itself to “+1 = -1” (and this is the first error (1 of 4) discussed in my preprint)

Best,

Florin

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Joy Christian replied on Sep. 5, 2011 @ 13:57 GMT
Dear Florin,

I am sorry, but this is nonsense. I came back from the FQXi conference last week where I discussed my equations (3) and (4) with David Hestenes. He understood exactly what I was talking about. He did not confuse theses two equations. He understood that one defines a fixed bivector basis and the other defines a fluctuating bivector basis. You, on the other hand, do not seem to have understood my model at all. This is very surprising to me, because until now I thought that you were one of the few people who did understand my model. In any case, the above equations are two completely separate equations, one involving a random variable and other being devoid of the random variable. The two do not have to match at all the time, because they both play completely different physical and statistical role in my model. They only have to agree at the point of actual measurement result, giving either +1 or -1 as a result. And that is exactly what the equations are doing. As I have said many times before, your confusions arise because you have been relying on my older papers, which are formulated with the standard scores (i.e., bivectors---mere mathematical constructs), whereas the paper we are discussing now is based on both the raw scores and the standard scores and the correct statistical relationship between the two. You are bound to run into inconsistencies if you mix things up haphazardly. If you honestly want to understand my model you would read the Eberlein paper, and then my latest paper, and then learn about the fact that standard scores are a mere mathematical constructs, not something that is actually observed in the experiments. What is actually observed in the experiments exactly matches with what is predicted by my model. There are no inconsistencies in my model.

Joy

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T H Ray wrote on Sep. 5, 2011 @ 12:38 GMT
Florin,

I'm afraid that any purported falsification of Joy Christian's local realistic framework by computer simulation is going to suffer the same flaw that Christian himself introduced in claiming "disproof" of Bell's theorem, which is the only part of Christian's argument that I reject.

The flaw is in assuming what one sets out to prove. Any way one slices it, "disproving" a...

view entire post


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Joy Christian replied on Sep. 5, 2011 @ 13:00 GMT
Dear Florin,

My equation (4) is not incorrect. It is a postulate, not a deduction from equation (3). A postulate cannot be incorrect. The motivation behind it comes from the already existing sign ambiguity discussed by Eberlein for example (cited in some of my papers). I strongly urge you to study this paper to recognize where you have gone wrong. You have got your mathematics all wrong, because you have not understood what the hidden variable in my model is. It is the handedness of the basis bivectors, or the orientation of the 3-sphere. This orientation "fluctuates" from right to left and back.

There is an easy way to bring out the running error in your paper. Whatever notations and conventions are used, the following two constraints on the basis bivectors must be satisfied:

For the right-handed basis bivectors one must maintain

(B_i)(B_j)(B_k) = +1 .

And for the left-handed basis bivectors one must maintain

(B_i)(B_j)(B_k) = -1 .

The hidden variable is then these two alternative possibilities.

If you maintain these constrains as consistency conditions throughout, then you will not be confused as you have been. And you certainly have been. Please read the Eberlein paper thoroughly.

Now, as you can easily check, the above two constraints are rigorously maintained in all of my papers, whatever the notations used.

On the other hand, these constraints are not satisfied by some of the crucial equations in your paper. Thus, I am afraid, your whole analysis is fundamentally flawed. By posting your fallacious analysis on the arXiv you have made my already difficult task much more difficult. Sadly, you are not helping physics by doing this. You are killing it off prematurely.

Best,

Joy

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Florin Moldoveanu replied on Sep. 5, 2011 @ 13:19 GMT
Dear Joy,

You can certainly have the freedom to choose any postulates you want, provided you do not choose inconsistent ones (because you can then derive any statement and its negation at the same time). As a postulate, Eq. 4 is inconsistent with Eq. 3 (Eq. 3 it is true). LHS of Eq. 3 is equal with LHS of Eq. 4 and therefore their right hand sides should be equal as well. Equating the RHS of Eq. 3 and 4 and simplifying the Kroneker delta and the Levi Civita results in: lambda = 1 in contradiction with your lambda = plus and minus one. In fact the inconsistency reduces itself to “+1 = -1” (and this is the first error (1 of 4) discussed in my preprint)

Best,

Florin

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T H Ray replied on Sep. 5, 2011 @ 13:42 GMT
Oh come on, Florin. The notation (which should actually be + 1 in one case and - 1 in another) is careless, but not wrong. It merely reflects a vanishing identity in the transitivity of the simple algebra, 1 - 1 = 0.

Tom

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Florin Moldoveanu wrote on Sep. 5, 2011 @ 13:01 GMT
Dear Tom,

You say: "I'm afraid that any purported falsification of Joy Christian's local realistic framework by computer simulation is going to suffer the same flaw that Christian himself introduced in claiming "disproof" of Bell's theorem, which is the only part of Christian's argument that I reject."

Please read my archive paper. I am not basing the falsification of the model on computer simulation. This was only how the investigation started. My rejection is based on mathematical mistakes within the model itself. They were a really big surprise for me as I did not expect them there and I had publicly stated before that I believe Joy's math to be correct and I only disagreed on his interpretation.

Computer simulation is very demanding though and any modeling mistakes would result in standard Bell's correlation and you have to be very careful. It was only after achieving Tsirelson's bound of 2 sqrt(2) with other models that boosted the confidence that Joy's theory should get to the 2 str(2) as well. There was a straightforward almost mechanical process of translating Joy's model into computer code and this process uncovered the errors. And the more I dug, the more mathematical mistakes I found and after reaching 4, I stopped.

The mistakes were found by good old fashion paper and pencil, and not by computer simulation (and in fact I never completed the computer simulation of Joy's model). Please read my paper for the explanation of the mistakes. They get progressively harder, but the first one is as simple as a misplaced minus sign. Unfortunately, this occurs at a critical point and it is not a typo you can recover from by a later change of the model. Fixing this error can be done only at the expense of breaking other things which are just as important to correctly model experimental results.

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Florin Moldoveanu wrote on Sep. 5, 2011 @ 14:23 GMT
Dear Joy,

I enjoyed this rapid exchanges allowed by our narrow time slice window when we are both on-line due to the time zone differences. I am now going off line and I’ll be back later with answers to any more questions from you or anybody else.

Best,

Florin

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Ray Munroe wrote on Sep. 5, 2011 @ 15:13 GMT
I agree with Florin that a vector cross-product is a pseudo-vector (also called an axial vector) with odd parity symmetry. It is simple quaternion algebra - nothing that can be confused and cloaked in more complex mathematics.

I think this gets to the heart of covariant and contravariant operators as I noted earlier. I also thought there were problems earlier in the derivation. Florin acknowledges that this is hyperbolic geometry, and I insist that complex number representations may emerge from improper mixing of the 'real space-like hypersurface' and the 'hyperbolic time-like radial' components of this geometry. But, sadly, no one else has reinforced my position.

Have Fun!

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Joy Christian replied on Sep. 5, 2011 @ 16:36 GMT
Ray,

I too agree with Florin that "a vector cross-product is a pseudo-vector (also called an axial vector) with odd parity symmetry" (see, e.g., my first paper).

Where I differ from Florin is where he has misinterpreted my equations and what they are supposed to represent in the model. He refuses to address my repeated call that he reads my latest two preprints. That would completely undermine his entire argument which derives from the earlier outdated criticisms by others. More to the point, consistency check of his claims have convinced me that he has no case against my model. The case he does have is against some model that he thinks is mine.

Joy

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Florin Moldoveanu replied on Sep. 5, 2011 @ 18:40 GMT
Dear Ray,

Indeed, please see my Eq. 4 and the discussion about Eqs. 23 and 24 on http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.1958v1.pdf In there Eq 24 is wrong as a X b should gain an additional minus sign when changing handedness because a X b is a PSEUDO-vector.

Dear Joy,

I had read carefully all your papers, including your latest two preprints. Every single one of them has mathematical mistakes, and in fact you challenge me earlier to point out mistakes in the one next to last preprint (http://arxiv.org/PS_cache/arxiv/pdf/1103/1103.1879v1.pdf) which I did. Also if you carefully read my preprint: http://arxiv.org/PS_cache/arxiv/pdf/1109/1109.535v1.pdf you notice that errors 3 and 4: illegal limit, and wrong rotor direction are in fact about your latest preprint: http://arxiv.org/PS_cache/arxiv/pdf/1106/1106.0748v3.pdf

Therefore your assertion that: “He refuses to address my repeated call that he reads my latest two preprints. That would completely undermine his entire argument which derives from the earlier outdated criticisms by others.” is patently false.

Best,

Florin

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Joy Christian replied on Sep. 5, 2011 @ 19:14 GMT
Dear Florin,

I am pleased to note that you have read my latest preprints. However by now I have spent some time on your arguments and, I am sorry to say, all of them are either wrong or unjustified. Apart from technical erros, the main reason they are wrong is that you have made up your own model which has nothing to do with mine. Equations (3) and (4) we discussed before are playing very different physical roles in my model. One produces bivectors that depend on the initial condition of the EPR pair (i.e., on lambda), and the other produces bivectors that are measuring devices independent of the initial condition lambda. This is much better explained in my latest paper which you are saying you have read. You cannot identify the two sides of equations (3) and (4) haphazardly as you have done. If you do that, then you will get nonsense, and that is what you are getting. You really have not understood my model after all. The very fact that you tried to simulate it, after all the discussions we had about the simulation, should have told us that you haven't really understood my argument. I know that you are determined to defend the orthodoxy, but you cannot do that by creating your own straw-man to knock down. You have to pay closer attention to what my model is actually saying. It is saying something completely different from what you claim it is saying. You have found erros in your straw-man, not in my model. In my model there are no errors.

Best,

Joy

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Steve Dufourny wrote on Sep. 5, 2011 @ 17:27 GMT
Sometimes I say me how many people really understands what is the geometric algebra.

Experts you say in geometric algebras and relativity. Well well well we have not the same definition of what is an expert. Expert in business strategy , that's all.

Steve and come on APS linkedin

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Steve Dufourny wrote on Sep. 5, 2011 @ 17:36 GMT
In fact neither Joy, nor Lisi nor Tom Or Ray really understands the foundamentals. It is simple in fact the truth you know.

Where is the evolution?, the thermodynamics?, the BE statistics?,the real convergences in 3d, the form of particles....all that is a pure joke...

Steve

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R. Onestone wrote on Sep. 6, 2011 @ 05:57 GMT
Mr. Christian,

"With sadness, I see that you have continued to ignore my latest papers and depended on the older, outdated debates."

Can a scientific article be too old, if it is correct? If some of your articles are deprecated, you should add a short note to each one of them on the arxiv.

"I do not agree with the so-called mistakes in my paper you mention"

By reading Dr. Moldovenau's paper I could see that the mistakes he pointed are from your most recent work. Since you claimed the opposite, your first answer was obviously given before reading his article. Did you reject his arguments before reading them?

Mathematics should be correct, no matter what physical interpretation you gave to it. A sequence of normed quaternions cannot have as limit 0. A unit pseudoscalar cannot be equal to its opposite. These are elementary mathematical facts. Dr. Moldovenau did not deduce them from a straw man model but they are written in your paper.

Rob

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Lawrence B. Crowell wrote on Sep. 6, 2011 @ 13:58 GMT
Florin points out that with β_j = β_j(λ) that

β_iβ_j(λ) = λ β_i β_j = -λ(δ_{ij} + ε_{ijk}β_k)

which leads to a problem with the value of λ. This is sort of Lie algebraic version of the same problem that has existed from the beginning. The construction has problems, where if β_i(λ) is Lie algebraic, then the product

β_i(λ)β_j(λ) = -δ_{ij} - ε_{ijk}β_k(λ)

= -δ_{ij} - λε_{ijk}β_k,

However by definition β_i(λ) = λβ_i we also have

β_i(λ)β_j(λ) = λ^2β_iβ_j =

β_iβ_j = -δ_{ij} - ε_{ijk}β_k.

This is inconsistent if λ = -1.

This 1 = -1 problem gets buried away in more sophisticated forms, and this process appears to have no end. One can bury an error of this sort in ever more complex symbolic structures. The whole construction is simply inconsistent, and this inconsistency keeps being covered over with layers of auxiliary assumptions and constructions.

Cheers LC

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Joy Christian replied on Sep. 6, 2011 @ 14:55 GMT
Where on earth did the first equation of yours come from? It is silly. Nowhere in my papers do I make such identification. The rest of your argument is therefore irrelevant. There are no inconsistencies in my papers.

Joy

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T H Ray replied on Sep. 6, 2011 @ 14:58 GMT
Lawrence,

This is just not so. The assumption that the lambdas are equivalent is wrong and has always been wrong. It's a straw man assumption.

Translated to integrals -- and I have been repeatedly pointing this out with apparently no success at getting Joy Christian's critics to understand it -- the Christian framework is Lebesgue integrable (real analytical) on orthogonal axes. Meaning that results - 1 OR + 1 describe correlated but opposite properties on a single axis, with random variables on the orthogonal axis, and a null result from rotation through a 4 pi full time interval.

All of this SHOULD be clear from the meaning of "coordinate-free" correlation of properties (Hestenes) and it is becoming just as disappointing to me as to Joy that this meaning is misunderstood. One cannot COMPUTE correlations free of coordinates; one can, however, integrate functions of a coordinate free system of geometric correlations. The whole point is covered in my previous post quoting Einstein -- that "... it follows that the above assumption for tracts must also hold good for intervals of clock-time in the theory of relativity. Consequently it may be formulated as follows: -- if two identical clocks are going at the same rate at any time and at any place (being then in immediate proximity to each other), they will always go at the same rate, no matter where and when they are again compared with each other at one place."

Integrating over this interval of clocks rotating in opposite directions produces a strict correlation of properties " ... no matter where and when they are again compared ..." which means that the axis is not rigid, but depends only on the ORIENTATION of these properties by the topology and the initial condition by which the properties were found to be correlated. The most common mistake most make is to assume that time intervals apply to the clock -- which in relativity only describes the rate of physical processes -- while the true relativistic case of correlated clocks refers to time-reverse symmetry: clocks running at identical rates in opposite directions. These correlated rates correspond to geometric correlation of properties.

Nowhere does Joy Christian make the mistake of implying - 1 = 1.

Tom

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Lawrence B. Crowell replied on Sep. 6, 2011 @ 17:49 GMT
CJ,

The first equation comes right from the algebraic identity in your equation 3,

β_iβ_j = -δ_{ij} - ε_{ijk}β_k.

Now if I let β_j -- > β_j(λ) = λβ_j then

β_iβ_j(λ) = λβ_i β_j

and now I substitute in the algebraic identity

β_iβ_j(λ) = λβ_i β_j = -λ(δ_{ij} + ε_{ijk}β_k).

I then compute β_i(λ)β_j(λ) and the λ^2 just gives the algebraic identity in the second line in this post. The I do it with β_i(λ) = λβ_i and then without this identification with

β_i(λ)β_j(λ) = -δ_{ij} - ε_{ijk}β_k(λ)

=-δ_{ij} - λε_{ijk}β_k

For λ = -1 it is clear this is an inconsistency.

Tom,

I suppose I am not following your reasoning. What I have done is pretty elementary. λ is just a scalar. What I did follows from elementary rules of algebra, and the set up by JC.

Cheers LC

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Sridattadev wrote on Sep. 6, 2011 @ 14:29 GMT
Dear All,

All numbers are mathematically "absolutely" equal. Please see the proof of

zero = i = infinity and true meaning of singularity.

relatively everything = absolutely nothing

Love,

Sridattadev.

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Florin Moldoveanu wrote on Sep. 6, 2011 @ 15:33 GMT
Joy,

You state: “Tom,

I agree with your last point. But what Florin has been doing is extraordinarily damaging sociologically. So I must deal with it first. We cannot leave this dangling.

Joy ”

Now this is the very problem. Let’s concentrate on the mathematical validity of the claims first and postpone the sociological implications later. I have yet to see a single mathematical argument against my claims and instead you kept focusing on: “ignore my latest papers”, “I am sorry, but this is nonsense. I came back from the FQXi conference last week where I discussed my equations (3) and (4) with David Hestenes”, “Now I am becoming suspicious of your understanding of geometric algebra as well”, “Apart from technical erros, the main reason they are wrong is that you have made up your own model which has nothing to do with mine.”, etc, etc

Those are all fallacious arguments and to my repeated specific citations of the errors in your papers down the actual equation or equation lines I have yet to see a single MATHEMATICAL rebuttal.

Best,

Florin

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T H Ray replied on Sep. 6, 2011 @ 15:57 GMT
Egad. Florin, if you find my rebuttal to your counterarguments "unmathematical," surely you cannot say it of Joy's very clear reply: "It is not all that mysterious to see if you really want to see it. Just separate out the two possible equations that are combined in equation (4) and see what they really are. One, with lambda = +1, is the even subalgebra of right-handed bivectors, and the other, with lambda = -1, is the even subalgebra of left-handed bivectors (originally proposed by Hamilton). And as I have mentioned before, for the right-handed bivectors we would then have (B_i)(B_j)(B_k) = +1 , and for the left-handed bivectors we would have (B_i)(B_j)(B_k) = -1 . It is as simple as that. I see no inconsistency whatsoever of any kind, anywhere." (5 Sept @20:40 GMT)

You see "coordinate frame" mentioned here?

Tom

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Joy Christian replied on Sep. 6, 2011 @ 16:00 GMT
Florin,

You know very well that to counter any arguments like the ones you have made---no matter how fallacious and misguided they are---requires time and effort. In a due course I will counter your arguments mathematically on the arXiv.

In any case, you HAVE ignored my latest papers, and I HAVE given you clear mathematical arguments on these very pages. Just look up my previous posts here. And I keep saying that you have ignored my latest papers because I see no hint of the distinction between raw scores and standard scores etc. etc. in any of your remarks, here or elsewhere. So, even if you did read my latest paper, you must have read it quite superficially. Otherewise you would not have made some of the claims you have made.

Best,

Joy

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Florin Moldoveanu replied on Sep. 7, 2011 @ 01:35 GMT
Dear Joy,

You state: "You know very well that to counter any arguments like the ones you have made---no matter how fallacious and misguided they are---requires time and effort. In a due course I will counter your arguments mathematically on the arXiv."

I am looking forward at your arXiv reply, but you are again misleading in your statements and you know it. You were very quick to point out the mixing of the fixed and dynamic betas in Lawrence’s argument and you did not invoke the arXiv argument there. In this blog, my arguments are just as simple. In fact I had proven to you in two posts below that an average high school student can validate them.

All you have to prove me wrong is that the following critical equality holds in a left handed frame

“a ∧ b = −I (a × b)”

However, this equality is mathematically false in any frame, left handed or right handed.

I am willing to bet with you the following thing: you will never write any reply on arXiv rebutting my paper. Why? Because you already know I am right and just don’t want to admit it in order to save face.

Admitting you made a mathematical mistake is hard, and when it involves your entire research program is very hard. Nobody is perfect and we all make mistakes. I too got blinded by the forest and did not see the trees when I stated publicly before that your model is mathematically correct and I only disagreed with your interpretation. My prior paper commenting on your work is now proven partially incorrect and I have to change it.

Best,

Florin

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Florin Moldoveanu wrote on Sep. 6, 2011 @ 18:43 GMT
Joy and Tom,

I have said earlier: “The fact remains that the so-called left handed identity: “a ∧ b := −I • (a × b).” from http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.1958v1.pdf is false even in a left handed coordinate frame. It is an elementary mathematical mistake and any average high school student with a half an hour introduction to geometric algebra can prove it as a homework problem.”

I am standing by this and I will actually do both the small tutorial and the homework problem in this and the next post to show it is really a trivial elementary mistake as I had claimed.

½ hour Geometric algebra tutorial:

Prerequisite: elementary vector algebra

Let a vector A be represented in a Cartesian coordinate as A = a1 e1 + a2 e2 + a3 e3 where e1,2,3 are the unit vectors, and a1,2,3 are the components of the vector.

In geometric algebra one gets to multiply vectors in an unlimited fashion provided that the order of the unit vectors is preserved. There are also two simple rules to simplify complicated expressions.

Rule 1: e1 e1 = e2 e2 = e3 e3 = 1

Rule 2: e1 e2 = -e2 e1 ; e1 e3 = -e3 e1 ; e2 e3 = - e3 e2

Also vector multiplication follows this rule: A B = A dot B + A hat B where dot is the ordinary vector product and the hat product is the so-called exterior product. For vectors we have A dot B = (A B + B A)/2 and A hat B = (A B – B A)/2

For convenience we name e1 e2 e3 = I

Exercise: multiply and then simplify (2 e1 e3 + 5 + 4 e2) (2 e2)

Answer: 4 e1 e3 e2 + 10 e2 + 8 e2 e2 = - 4 e1 e2 e3 + 10 e2 +8 = -4 I + 10 e2 +8

Now the homework problem: compute “a ∧ b” and “I (a × b)” in a left handed basis and compare them.

Best,

Florin

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Florin Moldoveanu wrote on Sep. 6, 2011 @ 18:44 GMT
Joy and Tom,

Here is the answer to the homework problem above.

Let e1 e2 e3 be a right handed coordinate system. Let e1 point towards you away from the paper, e2 point to the right and e3 point up. In this case I = e1 e2 e3. Construct a left handed coordinate system by mirroring e2: I = e1 (-e2) e3. Draw the two coordinate systems on paper to convince yourself they are right and left handed. In the first case “move” e1 on e2 with your right hand and you get a rotation along e3. In the second case “move” e1 on -e2 with your left hand and get a rotation also on e3. The only difference between the two rotations is their sense.

Compute a ∧ b

a ∧ b = a b – a dot b = (a1 e1 + a2 (-e2) + a3 e3) (b1 e1 + b2 (-e2) + b3 e3 ) – a dot b=

e1 (-e2) a1 b2 + e1 e3 a1 b3 + (-e2) e1 a2 b1 + (-e2) e3 a2 b3 + e3 e1 a3 b1 + e3 (-e2) a3 b2

Compute I (a X b)

I = e1 (-e2) e3

a X b = det(e1 (-e2) e3;a1 a2 a3;b1 b2 b3)

I (a X b) = e1 (-e2) e3 e1 (a2 b3 – a3 b2) - e1 (-e2) e3 (-e2) (a1 b3 – a3 b1) + e1 (-e2) e3 e3 (a1 b2 – a2 b1)

= -e2 e3(a2 b3 – a3 b2) + e1 e3 (a1 b3 – a3 b1) – e1 e2 (a1 b2 – a2 b1)

Compare the two expressions term by term and see they are identical. Therefore a ∧ b = I (a X b) EVEN IN A LEFT HANDED BASE.

Best,

Florin

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T H Ray replied on Sep. 6, 2011 @ 19:01 GMT
Florin,

Fine. Except that nowhere does Joy identify handedness with a coordinate system. I thought my example of correlated but separated ninety degree angles, LH & RH, sharing a common upright, would make the point. I thought Joy's description of handedness for basis bivectors would make the point.

I guess not. I am still at a loss to understand how anyone finds this obscure.

Tom

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Florin Moldoveanu replied on Sep. 6, 2011 @ 19:38 GMT
Tom,

Please see http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.1958v1.pdf and the discussion around Eq 23, 24.s

The first line after Eq 24 reads "along with the left-handed duality relation a ∧ b := −I · (a × b)."

This is the mistake. THERE IS NO SUCH THING AS "a ∧ b := −I · (a × b)" on either left or right hand basis. Here is the bogus minus sign which allows Joy to write Eq. 25 and after averaging on a 50-50 random distribution of handedness (his hidden variable), make the "mu" term vanish.

The last line of Eq. 17 in http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703179v3.pdf is wrong and the average in Eq. 19 of the same paper is wrong as well. Therefore correctly computing correlations in Joy's model results in -a dot b + I (a X b) while experiments and quantum mechanics has only - a dot b

Joy's model is a hidden variable model, but it does not yield any agreement with experiments and cannot be called a disproof. The troubling minus sign mistake apears in other places as well, and Joy commits other illegal mathematical opperations to "fix it".

Florin

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T H Ray replied on Sep. 6, 2011 @ 20:44 GMT
Florin,

Did you read my outline of a proposed experiment for local realism by Joy Christan's criteria (attached)? He hasn't said whether he agrees with it or not, and maybe he doesn't. However, I do think it captures the reasons why I am convinced that his criteria not only predict the coin toss probability of Bell-Aspect correlations, but also explains why mu does disappear on one axis to be replaced by pair correlation on the orthogonal axis, as a result of a complete rotation through one time interval. It makes the difference between probabilistic functions on an arbitrary space, and continuous functions on a complete spacetime.

(I think this also addresses your points in the separate thread, 6 Sep @ 1948 GMT, so I will post this just once here. Let me know.)

Tom

attachments: 1_Local_Realism_Handout.pdf

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Florin Moldoveanu wrote on Sep. 6, 2011 @ 19:48 GMT
Tom,

Please see Joy's first paper: http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703179v3.pdf

Aro
und Eq 14 it reads

"Taking aboard this hint, let us then venture to replace

the polar vector  of Bell’s model with the unit trivector

μ = u ^ v ^ w = ±I  ± ex ^ ey ^ ez , (14)

which can be pictured as a parallelepiped of unit volume,

assembled by the vectors u, v, and w of finite lengths and

arbitrary directions, giving it an unspecified shape and

orientation."

Here is the key assumption he makes: "u, v, and w of finite lengths and

arbitrary directions" Then in 50% of the cases one gets a right handed coordinate system {u,v,w} and in the other 50% a left handed coordinate system {u,v,w}.:

"This allows us to quantify the ambivalence in the orientation of μ

simply by the sign of I. In what follows, we shall take the

vectors u, v, and w—and hence μ—to be uncontrollable"

Then he average the correlations on an oriented manifold V3 meaning averaging it on the two handedness. But the problem is that on both handedness, Hodge duality maintains its overall plus sign and the average does not eliminate the cross product term.

Florin

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Lawrence B. Crowell wrote on Sep. 6, 2011 @ 20:31 GMT
Tom,

I am starting anew here. Let me begin with

β_iβ_j = -δ_{ij} - ε_{ijk}β_k,

which is just a Lie algebraic statement if I commute i and j we get

β_jβ_i = -δ_{ji} - ε_{jik}β_k.

Now let me subtract these

β_iβ_j - β_jβ_i = - ε_{ijk}β_k - ε_{jik}β_k,

which in a more compact form and with ε_{jik}β_k = -ε_{ijk}β_k we have

[β_i, β_j] = -2ε_{jik}β_k.

This is a basic so(n) type of algebra. Now suppose I flip the orientation of β_i -- > -β_i. It pretty trivially follows that

-[β_i, β_j] = 2ε_{jik}β_k,

(negative sign on left, and ij flip on right) and we simply recover the same result.

This commutator defines the element of the 2-form

β/\β’ = [β_i, β_j]ω^i/\ω^j,

which is computed by Florin above with β/\β’ = βxβ’ - β*β’. Clearly the handedness of the elements has no contribution.

Another way to look at this is the commutator in general is

[β_i, β_j] = -2C_{jik}β_k,

where C_{ijk} are the invarant structure constant of the Lie algebra. The commutator is the adjoint representation of a Lie algebra

ad(x)(y) = [x, y], --- > or ad(x) = [x, _]

and the adjoint representation satisfies ad(λx) = λad(x) = λ[x, _]. So if I have ad(λx)(y) then

ad(λx)(y) = λad(x)(y) = λ[x, y]

and I can divide out the λ and see that the result is unchanged. I can go further with ad(λx)(λy) then clearly

ad(λx)(λy) = λad(x)ad(λy) = λ[x, λy] = λ^2[x, y].

Then for [x, y].= z z -- > λ^2z, and by ad(x)(y) = -ad(y)(x) it is clear that

ad(λx)(λy) = λ^2ad(x)ad(y) = λ^2[x, y].

and again I divide out by λ^2 and the structure constant is an invariant.

Now if I impose a condition that there is

β_iβ_j = -δ_{ij} - λε_{ijk}β_k,

I then run into a host of difficulties. In effect this amounts to saying the structure constants obey C_{ijk} = -C_{ijk}, which implies they are zero and this is an abelian group.

Cheers LC

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Florin Moldoveanu replied on Sep. 6, 2011 @ 20:40 GMT
Thanks Lawrence.

I think this concludes the discussion on error 1 unless Joy or Tom objects.

The next issue is that: is this error recoverable? Can one modify Joy's model in any way to fix this problem? Sadly it is not possible. It would have been much more interesting if the answer was yes.

Florin

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T H Ray replied on Sep. 6, 2011 @ 20:54 GMT
I don't know when Lie algebra entered this discussion, but it certainly does not correspond to Joy's bivector basis.

Tom

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Lawrence B. Crowell replied on Sep. 6, 2011 @ 21:25 GMT
Florin,

The result that C_{ijk} = -C_{ijk} leads to an abelian group seems to play into Bell’s result and give a justification for the classical commutativity. This seems to demolish the classical non-commutativity with orientation dependent λ of JC’s model. Maybe this leads to some Lie algebraic variant of a Bell type of theorem.

Tom,

Yes, Lie algebras do enter into the picture. The orientation logic of vectors in 3-dim involves the SO(3) group, with the [β_i, β_j] = ε_{jik}β_k algebra for the group generators.

Cheers LC

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Lawrence B. Crowell wrote on Sep. 7, 2011 @ 02:56 GMT
I agree with citing of error #2. Equation 19 of arXiv:quant-ph/0703179

∫(μ*a)( μ*b)dρ(μ) = ∫a*bdρ(μ) - ∫μ(axb)dρ(μ) = a*b + 0

means that a/\b = I*(axb) is such that when integrated over the volume I = I(e1, e2, e3) is zero. I(e1, e2, e3) is a rhomboid or parallelepiped, and the above integral has μ = I. This leads to the 1 = -1 problem again, for this assumes the orientation of the

I(e1, e2, e3) = e1/\e2/\e3

changes in the integration due to this handedness issue. Since that is false the second integral is not in general zero. This integral should read

∫(μ*a)( μ*b)dρ(μ) = ∫a*bdρ(μ) - ∫I*(axb)dρ(μ) = a*b + - ∫I*(axb)dρ(μ).

Cheers LC

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Florin Moldoveanu replied on Sep. 7, 2011 @ 04:09 GMT
Lawrence,

Error # 2 is about the impossibility of changing anything in the Clifford algebra formalism to save Joy Christian's hidden variable model. Fixing the +1=-1 mistake in the correlation computation (by any hypotetical modification of the model) breaks the local average for Alice or Bob. You need them both however to get agreement with experimental results.

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Lawrence B. Crowell replied on Sep. 7, 2011 @ 15:27 GMT
The interesting thing about all of this is it seems to be pointing to some aspect to the foundation of QM. I have not read Clifton'spaper yet, though I downloaded it yesterday. JC's work, even with its problems, seems to point to some aspect to q-foundations.

I am trying to keep things brief today --- I spent a bit too much time on this yesterday.

Cheers LC

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Joy Christian replied on Sep. 7, 2011 @ 16:01 GMT
There is no handedness or any other kind of problem in my model. Neither Florin nor you have understood the notation I have used in my first paper, and hence your equations above have nothing much to do with my model. Even after I have explained what the notation means in some nine papers, Florin has managed to eagerly confuse himself and others about my notation. There is nothing wrong with any of the equations in any of my papers. The errors lie in Florin's own understanding of my notation, and more surprisingly in his own understanding of basic geometric algebra.

Joy

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Joy Christian wrote on Sep. 7, 2011 @ 08:21 GMT
Dear James,

Let me try to address a neutral person (you) for reasons that will become clear soon.

I see Florin gloating and self-congratulating on these pages; but for what? He continues to triumphantly refer to "error 1", "error 2", etc., as if they are actual errors in my work, not his misconceptions. In fact they are a result of his own errors, as I have tried repeatedly to...

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T H Ray wrote on Sep. 7, 2011 @ 10:07 GMT
Joy,

This is as straightforward and clear an explanation as I think anyone is capable of giving.

I am now convinced that factors other than mathematics are driving this dialogue, and I'm beginning to appreciate the magnitude of what you have dealt with in the past four years.

Tom

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Joy Christian replied on Sep. 7, 2011 @ 10:23 GMT
Thanks, Tom.

Joy

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Peter Jackson wrote on Sep. 7, 2011 @ 10:38 GMT
Joy

I appreciated your clarification to James. I've found many consistencies in the results of your brilliant thinking. Unfortunately one of the great strengths of humankind, our disparity, also derives our greatest weakness, failure of comprehension from other viewpoints. As and Einstein said Bragg said, we have to learn how to look and 'think' and differently if we are to understand. Many cannot do so, as we amply find here.

The greater reason I have full confidence in your work is that I have tested it by arriving at the same place by an entirely different route, conceptual and empirical, so it derives a fully falsifiable, ontological and predictive Local Reality.

I would however, to be fair, like to see if you are capable of achieving what you are asking Florin to achieve. It may be a greater task than you realise.

A few weeks ago you undertook to read and respond on the re-interpretations in my finalist essay; (20:20 Vision) http://fqxi.org/community/forum/topic/803

You may well have done and not understood/considered it nonsense. It is indeed already well 'out of date'. But I hope you may demonstrate that you can do what you expect Florin to do to properly review from your own viewpoint. Take 5 steps backwards to allow a genuine conceptual overview and re-appraisal 'of that which nature has revealed to us'.

Other joint papers are on the way, but slightly fuller earlier re-appraisal was published in the GSJ and posted at; http://wbabin.net/weuro/jackson.pdf

The model that worked does not represent any 'new theory' but is simply a completion of the jigsaw puzzle with the pieces we have, and the final picture is entirely consistent with your mathematical proof. I'd be grateful of your comments, but also hope you may find great value from the fresh route found from Bragg & AE's advice.

James - I hope you're also progressing. you may also be interested in the most recent update to the GSJ paper, including deriving Stellar Aberration direct from waves at zero crossing speed (perpendiclar) incidence, which also explains the 'correction factors', and inconsistencies, not understood since Kimura over 100 years ago! If you're unfamiliar with the concept of 'Proper Time', as most are, familiarising yourself with it first should help.

Very best regards.

Peter

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Joy Christian replied on Sep. 7, 2011 @ 12:28 GMT
Peter,

Many thanks for your encouraging words. I did promise to read your essay, but last two weeks went by with the FQXi conference, and now when I am back there seems to be endless things I must attend to. This is not an excuse, but truth. One of these things is to write a comprehensive response to Florin's criticism of my work that has just appeared on the arXiv. His arguments are all wrong and naive, but if I do not respond to them then they will be taken to be true by many. So, I am afraid, that will have to be my first priority. But I will try to read your essay at some point.

Many thanks again,

Joy

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Florin Moldoveanu wrote on Sep. 7, 2011 @ 18:41 GMT
Dear Joy,

I offer you a bet (again): You will not write any rebuttal on the archive against my preprint because you know I am right and just don’t want to admit it to save face. It is usually said: the cover-up is worse than the offense.

I am not gloating, I am actually sad seeing you on the path to self-destruction with insults and lies instead of accepting a high school level mistake.

So here is my last direct question to you. In http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.1958v1.pdf you have the so-called left handed identity: “a ∧ b = −I (a × b)” in a left handed frame. This is at the core of mistake number one which I present in my preprint. It is very early in you derivations and ruins everything else after it making a moot point to discuss anything else like your raw scores.

Please accept this identity is false and your hidden variable model does not reproduce any experimental results, OR please provide a proof that the identity is right. In the next post I will provide an elementary proof that in a left handed base the correct identity is “a ∧ b = +I (a × b)”. This identity is nothing but the Hodge duality mapping bivectors to axial vectors (http://en.wikipedia.org/wiki/Hodge_dual).

Best,

Florin

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T H Ray replied on Sep. 7, 2011 @ 20:18 GMT
Florin,

Do you really not understand that handedness can be expressed in terms of complete objects sharing a common boundary, rather than in the transitivity and reflexivity of equal quantities in an algebraic equation?

Tom

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Florin Moldoveanu replied on Sep. 7, 2011 @ 21:23 GMT
Tom,

You state: "handedness can be expressed in terms of complete objects sharing a common boundary?"

Sorry but this is just meaningless. Pick up a physical right hand screw and and a left hand screw from any hardware store. They are a "complete object"(s) are they not? (yes) What boundary do they share? (none) Can you tell their handednes appart just by looking at them one at a time? (yes)

Please see http://en.wikipedia.org/wiki/Right-hand_rule for clarifications on handedness.

Florin

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T H Ray replied on Sep. 8, 2011 @ 08:43 GMT
Florin,

Actually, it's your physical analogy, and not my mathematical statement, that's meaningless.

Recall that it is just this "at a time" comparison that you make with the screws, that I first realized as the problem with Bell-Aspect measurements, as I deconstructed Joy's explanation.

Joy's framework incorporates continuous spacetime consistent with EPR and with hidden variables. The transformation from LH to RH is a continuous operation -- not dual, not statistical, not discontinuous in any way. It results from the topological orientation of a fully integrated spacetime measure over an arbitrary time interval. Thus, the remarkable result is that what we measure is real analytical (Lebesgue integrable - oo -- + oo) supporting Einstein's and Minkowski's assumption that spacetime -- and neither space nor time taken separately -- is physically real.

I will have more to say.

Tom

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Florin Moldoveanu wrote on Sep. 7, 2011 @ 18:42 GMT
Elementary proof that there is no sign change in the Hodge duality mapping bivectors to axial vectors in a left handed frame.

Let e1 e2 e3 be a right handed coordinate system. Let e1 point towards you away from the paper, e2 point to the right and e3 point up. In this case I = e1 e2 e3. Construct a left handed coordinate system by mirroring e2: I = e1 (-e2) e3. Draw the two coordinate systems on paper to convince yourself they are right and left handed. In the first case “move” e1 on e2 with your right hand and you get a rotation along e3. In the second case “move” e1 on -e2 with your left hand and get a rotation also on e3. The only difference between the two rotations is their sense.

Compute a ∧ b

a ∧ b = a b – a dot b = (a1 e1 + a2 (-e2) + a3 e3) (b1 e1 + b2 (-e2) + b3 e3 ) – a dot b=

e1 (-e2) a1 b2 + e1 e3 a1 b3 + (-e2) e1 a2 b1 + (-e2) e3 a2 b3 + e3 e1 a3 b1 + e3 (-e2) a3 b2

Compute I (a X b)

I = e1 (-e2) e3

a X b = det(e1 (-e2) e3;a1 a2 a3;b1 b2 b3)

I (a X b) = e1 (-e2) e3 e1 (a2 b3 – a3 b2) - e1 (-e2) e3 (-e2) (a1 b3 – a3 b1) + e1 (-e2) e3 e3 (a1 b2 – a2 b1)

= -e2 e3(a2 b3 – a3 b2) + e1 e3 (a1 b3 – a3 b1) – e1 e2 (a1 b2 – a2 b1)

Compare the two expressions term by term and see they are identical. Therefore a ∧ b = I (a X b) EVEN IN A LEFT HANDED BASE.

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T H Ray replied on Sep. 7, 2011 @ 20:20 GMT
There IS NO preferred coordinate system in Joy Christian's framework. To assign one, makes no sense.

Tom

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Florin Moldoveanu replied on Sep. 7, 2011 @ 21:14 GMT
Tom,

I beg to differ. Quoting from Joy's first paper: http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703179v3.pdf

"Ta
king aboard this hint, let us then venture to replace

the polar vector  of Bell’s model with the unit trivector

μ = u ^ v ^ w = ±I  ± ex ^ ey ^ ez , (14)

which can be pictured as a parallelepiped of unit volume,

assembled by the vectors u, v, and w of finite lengths and

arbitrary directions, giving it an unspecified shape and

orientation. Here the second of the equalities follows from

the fact that every trivector in the algebra Cl3,0 differs

from I only by its volume and orientation. This allows

us to quantify the ambivalence in the orientation of μ

simply by the sign of I."

His hidden variable MU is the heandedness of an arbitrary coordinate system made by arbitrary picking vectors u, v, and w of finite lengths and

arbitrary directions. Then ideed he is rigth stating: "quantify the ambivalence in the orientation of μ simply by the sign of I."

Then Joy proceeds to compute a correlation ON ALL random handedness on what he calls "a smooth orientable vector manifold V3". There he uses the indefinite duality relationship ("and duality relation a ^ b = μ(a × b) have been used.") made by the combining the right handed Hodge duality with a hypothetical left handed duality. In the right handed coordinate systems his indefinite relationship reads: "a ^ b = + I(a × b)" which is correct. On the left handed coordinate systems his indefinite relationship reads: "a ^ b = - I(a × b)" which is incorrect as shown above.

The correct relationship is "a ^ b = + I(a × b)" in both cases on V3 instead of "a ^ b = μ(a × b)" and this kills a MU factor in integral 19 in his paper. Therefore his final answer cannot be -a dot b but should be -a dot b - a hat b instead (and this contradicts experimental results). Please see the "+ 0" in the last line of Eq. 19. That is incorrect, it should be "- a hat b".

The reason Joy missed a sign is that in left handed coordinates I becomes -I as he correctly writes, but (a X b) becomes -(a X b) as well and he missed that. Both "I" and "(a X b)" are PSEUDO-objects changing signs on changing handedness. Joy simply forgot (a X b) is a PSEUDO-vector, that's all.

Florin

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Florin Moldoveanu replied on Sep. 7, 2011 @ 21:35 GMT
Tom,

You state: "There IS NO preferred coordinate system in Joy Christian's framework. To assign one, makes no sense."

I think I did not reply directly to your statement in my prior answer. Let me rectify this now.

Indeed, there IS NO preferred coordinate system. However, Joy computes 2 averages (Eq 18 and 19 in his first paper) on ALL coordinate systems. To proceed with his computations he separates the right handed and left handed coordinate systems and applies different duality relationships on them to prove the elimination of the the unphysical "-a hat b" term which he writes after integration as "+0" in Eq. 19. Trouble is that the left handed duality in incorrect and "-a hat b" does not vanish.

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Florin Moldoveanu wrote on Sep. 8, 2011 @ 02:27 GMT
Hello everybody. I need your help in making the proof below as easy and as straightforward as possible. Please let me know if there is anything that needs further explaining. Thank you!

Elementary proof that there is no sign change in the Hodge duality mapping bivectors to axial vectors in a left handed frame. If the proof below is correct, all Joy Christian’s results related to...

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Lawrence B. Crowell replied on Sep. 8, 2011 @ 02:50 GMT
I would say this looks good. There are a couple of "funnies." You have

WHY? There are 3 terms which simplifies: a1 b1 e1 e1 + a2 b2 (-e2)(0e2) + a3 b3 – a dot b

where the appearance of a 0 should be - isgn, an obvious typo. You also have

Therefore a ? b = I (a X b) EVEN IN A LEFT HANDED BASE.

where ? should be a wedge symbol

Cheers LC

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Florin Moldoveanu replied on Sep. 8, 2011 @ 03:57 GMT
Thanks Lawrence,

Indeed 0 and - are a typo, 0 is located on the keyboard next to -.

Also ? and hat got mixed up when I pasted in Word and I did not noticed copying it back to the small text editor.

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Jenny Harrison wrote on Sep. 8, 2011 @ 02:56 GMT
Dear Joy,

Can you tell us precisely why Florin's analysis does not apply to your work, and not refer us to other papers or other people's opinions? The mathematics seems straightforward in his last post, but perhaps his conclusion is not relevant for some reason which we can understand. Let us hope for your sake that either his proof is wrong or his conclusion is not relevant. I see nothing wrong from one step to the next. I continue to hope you are right because you have invested so much in this and I like you, but this concerns me.

Best regards,

Jenny

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Florin Moldoveanu replied on Sep. 8, 2011 @ 05:26 GMT
Dear Jenny,

I too invested very much time into understanding Joy’s model in the hope of contributing to this area. I must confess I was extremely disappointed when I first found out the mistakes and you have to believe me that I was not aware of them earlier at the time of posting part 1 of “To be or not to be (a local realist)” otherwise I would not have posted an archive paper stating in the abstract: “If Bell’s theorem’s importance is to rule out contextual hidden variable theories obeying relativistic locality, then Joy Christian’s counterexample achieves its aim.” which by now it is a proven false statement.

I will let Joy answer if the question of the validity of the math (which I repeatedly asked him to do instead of focusing on the sociological impact of the discovery). But I can show you why the sign error in the Hodge duality mapping bivectors to axial vectors in a left handed frames is critical and collapses all Joy’s claims of a disproof of Bell’s theorem.

Please view paper http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.1958v1.pdf first on page 7 around Eqs 23 and 24. Joy Combines the two equations on the two handedness into Eq. 25 with “the combined duality relation a ∧ b := μ • (a × b).”

Now please view his original paper http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703179v3.pdf and take a look at Eqs. 17, 18, 19

Eq. 17 is the same as Eq. 25 from the prior paper. Eq. 19 uses Eq. 17 to compute the average for Alice and Bob correlations resulting in – a dot b + 0. The trouble is with +0 which is incorrect. Anything else except +0 means his model predicts something else than experimental results. And it is not me saying it, it is Joy himself in his response to Marcin Powlowski (see Response 1 on http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703244v12.pdf ): “More specifically, the critics culminate their charge by declaring that, within my ocal realistic framework, it would be impossible to derive “a scalar in the RHS of the CHSH inequality. QED.”

If this were true, then it would certainly be a genuine worry.”

However the correct result is not +0 but “- a hat b” and the worry is indeed genuine.

The answer is not zero because the left handed Hodge duality just reinforces the right handed Hodge duality and does not cancel it out on his V3 manifold. The sign for the left handed Hodge duality is just wrong and that is why it is so critical to prove that the sign is +1 on left handed bases and not -1 as Joy claims.

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Joy Christian replied on Sep. 8, 2011 @ 08:44 GMT
Dear Jenny,

Good to hear from you, and thank you for your concern for me. By the way, I was hoping to see you at the FQXi conference but in the end you did not turn up.

I would not worry too much about Florin's arguments. He has read my papers very selectively, incorrectly, and with the intention to destruct, not construct. If you notice he always refers to something or other I said four years ago, but refuses to address things I have said on this very blog in the last few days. He is committed to orthodoxy much more than he cares to admit. In any case, I have a clear response for each point Florin is making, but I am not going to reveal my arguments on this blog just yet. I came back from the FQXi conference last week with a severe cold and flu, and yet I have started working on a robust response to Florin's arguments. To give you a sneak preview, some (only some) of his mathematical arguments are correct, but his interpretation of my hidden variable model is quite incorrect. Please have a little more patience and you will have my response in precise mathematical terms.

By the way, for me this is not about my ego or self-preservation. You know me sufficiently well to know that. For me it is about truth. And Florin has done a hell of a job distorting it. We certainly cannot have that stand.

Joy

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Florin Moldoveanu replied on Sep. 8, 2011 @ 21:42 GMT
Dear Jenny,

I would not hold my breath for a “robust response”. In fact I have a challenge to Joy: he will never ever publish anything on the archive rebutting my preprint because he knows I am right and he cannot argue with a black and white contradiction +1 = -1. Instead he lays in the groundwork for a later saving face excuse: “committed to orthodoxy”. In other words he will later say: I am alone against the “establishment”, I was defeated by dark forces bent on destruction who had no clue of geometric algebra and had never given me a fair chance by their out of context misinterpretation. Or some other nonsense along those lines.

On this very blog he stated: “what Florin has been doing is extraordinarily damaging sociologically. So I must deal with it first. We cannot leave this dangling” Please see my reply on Sept. 6, @ 15:33 GMT:

“Now this is the very problem. Let’s concentrate on the mathematical validity of the claims first and postpone the sociological implications later. I have yet to see a single mathematical argument against my claims and instead you kept focusing on: “ignore my latest papers”, “I am sorry, but this is nonsense. I came back from the FQXi conference last week where I discussed my equations (3) and (4) with David Hestenes”, “Now I am becoming suspicious of your understanding of geometric algebra as well”, “Apart from technical erros, the main reason they are wrong is that you have made up your own model which has nothing to do with mine.”, etc, etc

Those are all fallacious arguments and to my repeated specific citations of the errors in your papers down the actual equation or equation lines I have yet to see a single MATHEMATICAL rebuttal.”

Joy had many mathematical errors in his papers. What I can and will do is to keep focusing on the very first one proving wrong the “left-handed duality relation a ∧ b := −I • (a × b).” in as easy terms as humanly possible and on why this is critical for all his claims. When you start with “+1 = -1” you can prove anything like: sky is pink, Elvis is alive, and Bell theorem is disproven. It is a shame Joy chose the self-destruction path of denials, cover-ups, insults, and lies instead of accepting he was only human and made a mistake and starting anew with a different research program he is more than capable of doing.

Florin

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Peter Jackson wrote on Sep. 8, 2011 @ 11:09 GMT
Joy / Florin

There's something I don't understand about Local Reality and co-variance, and perhaps you could both give your considered views.

Take light travelling from one medium to another, say from water to a gas (air will do, so accelerating from ~140,000 to 186,000miles/sec. But the media are moving laterally at significant relative velocity v.

As we know, observing from the water, (frame K) the light 'path' appears 'dragged' by the air, but observing from the air (frame K'), the observed path is refracted back towards the 'normal' or the perpendicular to the interface plane. Let us assume we're considering waves, but they could be made up of photons, and/or vice versa. (If you find it easier to consider this with the light signal going the other way please also do so).

Firstly; 1). Is there not a Local Reality in each frame/medium where c (or c/n in this case) and the laws of physics equally apply (as SR).? If not, please explain.

Secondly; 2). Would the 'common boundary' not qualify as 'handed' (or any equivalent term you wish) in relation to any signal path and the media's relative vector?

Lastly; 3). Do we not observe signal 'polarity' as the information on direction of source our lenses pass to us? (as evidenced by Birefringence etc). In which case do you agree that all signals do not necessarily coincide with the normal to the Schrodinger sphere surface or (causal) light cone surface?

I believe there are some quite fundamental implications about our understanding involved.

Best regards

Peter

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Florin Moldoveanu replied on Sep. 8, 2011 @ 15:59 GMT
Dear Peter,

Can you please provide a diagram of your experiment? Is this similar with the Hoek experiment? http://www.conspiracyoflight.com/Hoek/Hoek_Experiment.html

If the light source is strong (many photons) then standard electromagnetism theory applies and all you get is a confirmation of relativity because relativity is derivable from Maxwell's equations (the speed of light propagation in vacuum is independent of the state of motion and Einstein started from here when he discovered relativity). QM enters the picture only when you are doing the experiment with a dim light source, one photon at a time. In this case you can describe the same thing using QM and it will agree 100% with standard classical electromagnetism.

I think the analasys in both cases is quite complicated on par with homework problems from the standard Jackson textbook which usually requires a graduate student to spend between 1 to 3 hours. I do not quite have the bandwith of time to do the analasys and present it here in a easy to understand form and I am forced to give only high level argments instead.

I do not expect however this problem to have any fundamental implications.

Your better bet is to ask Lawrence who is much more versed in relativity than I am and he may know of a shorter and easier solution.

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Eckard Blumschein replied on Sep. 8, 2011 @ 16:58 GMT
Dear Florin,

Peter did not perform experiments. He imagines light propagating in media like gas or water and accordingly subject to dragging. While there are renowned experts like Styrkov who are claiming having found evidence for dragging the hypothetic medium "aether" by the earth, they do not equate to my knowledge this hypothetic medium with optical media.

You reiterated what is often said: "relativity is derivable from Maxwell's equations (the speed of light propagation in vacuum is independent of the state of motion and Einstein started from here when he discovered relativity)"

I consider this incorrect. Actually, Einstein's second postulate adopts the earlier known and accepted constancy of the speed of light in empty space which was postulated by Maxwell and found to be in agreement with measured data. "independent of the state of motion" is imprecise. It does not pinpoint what moves. Einstein wrote independent from the motion of the emitter. This was correct but insufficient. The speed must be independent from the motion of the receiver too. Otherwise the same wave would have different speeds when it hits two targets with different relative to it speeds.

Surely, not just Maxwell's equations but all wave equations are not covariant under the Galilei transformation. In Case of Maxwell's equations the invariance resides in Faraday's induction term. Lacking covariance means, Galileo's relativity does not apply. The first postulate is a petitio principi, maybe a useful to some extent approximation, no discovery but rather a fabrication to be scrutinized anew as those are begging who signed the petition.

If I am wrong, please do not hesitate to demonstrate that.

Regards,

Eckard

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Florin Moldoveanu replied on Sep. 8, 2011 @ 17:58 GMT
Dear Eckard,

Indeed, Peter did not perform an experiment. I used the word experiment like in "gedanken experiment", but this is only parsing words; I hope my meaning was clear.

However I have a request. Can we stick to QM and realism in this forum and not start discussing relativity issues here? There will be plenty of other opportunities to do just that.

Thanks,

Florin

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 02:55 GMT
Hello everybody. I need your help in making the proof below as easy and as straightforward as possible. Please let me know if there is anything that needs further explaining. Thank you!

Elementary proof that there is no sign change in the Hodge duality mapping bivectors to axial vectors in a left handed frame. If the proof below is correct, all Joy Christian’s results related to...

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 02:56 GMT
Hello everybody. I need your help in making the explanation below as easy and as straightforward as possible. Please let me know if there is anything that needs further explaining. Thank you!

Elementary argument why the sign error in the Hodge duality mapping bivectors to axial vectors in a left handed frame destroys agreement with experimental data in Joy Christian’s hidden variable...

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 02:57 GMT
Hello everybody. I need your help to support the following appeal to Joy Christian. If you agree with it please reply to this threads stating you support it

Dear Joy Christian,

In the prior 2 posts two elementary arguments were presented about a mistake in the duality relation: a ^ b = μ(a × b) on a left handed frame when MU = -I and its fundamental implication for your hidden variable model’s ability to reproduce experimental results.

The presentations are straightforward and elementary. Please either point out any mistakes in the prior 2 posts, or admit that your hidden variable model is not viable and does not “disproof” in any way Bell’s theorem.

If you disagree that the two posts above do not represent a genuine mathematical proof against your disproof of Bell’s theorem, please explain it in easy to understand mathematical terms and do not take the self destruction road of mud sliding (For me it is about truth. And Florin has done a hell of a job distorting it.), lies (there are no coordinate frames in either geometric algebra or my model), insults (Now I am becoming suspicious of your understanding of geometric algebra as well.), and non-mathematical excuses (I have a clear response for each point Florin is making, but I am not going to reveal my arguments on this blog just yet.).

Sincerely,

Florin Moldoveanu and all the other people supporting this appeal with their replies on this thread.

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Lawrence B. Crowell replied on Sep. 9, 2011 @ 03:14 GMT
I support this. I think JC said that he is going to write another blog page here in the near future. So we will have to see what he comes up with.

Cheers LC

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Florin Moldoveanu replied on Sep. 9, 2011 @ 04:00 GMT
Thanks Lawrence,

I means a lot to me.

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Joy Christian replied on Sep. 9, 2011 @ 06:51 GMT
I wholeheartedly support the intellectual aspect of this appeal (although I have no idea why an appeal is needed). I have repeatedly said on this page in the last few days that I will produce a comprehensive response to the issues raised by the recent two preprints criticizing my work on Bell's theorem. However, I strongly disapprove the coercing tactics and the pompous, self-righteous, deceptive, and condescending language being used by the originator of this appeal. For this reason this will be my last entry on this page. My response will only appear on the arXiv. And for those who do not wish to remain in suspense, there is nothing wrong with any of my equations and their interpretations in any of my papers. In particular, there is nothing wrong with the Eq. (17) of my first paper and any of the terms involved in it. In fact, that equation and the meaning of all of its terms have been derived and explained in many different ways throughout my already existing nine papers on the subject. But I will derive it yet again, in a different way, starting from the first principles, explaining every term, and elaborating on every intermediate step of the derivation, with full references and other academic paraphernalia. However, as I said, since I do not wish to be bullied any further, this will be my last words on this page.

Joy Christian

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 04:02 GMT
Sorry, typo... Let me do it again.

Thanks Lawrence,

It means a lot to me.

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John wrote on Sep. 9, 2011 @ 08:16 GMT
Dear all,

Joy mentioned in one of his replies to Florin:

"It is not all that mysterious to see if you really want to see it. Just separate out the two possible equations that are combined in equation (4) and see what they really are. One, with lambda = +1, is the even subalgebra of right-handed bivectors, and the other, with lambda = -1, is the even subalgebra of left-handed bivectors (originally proposed by Hamilton). And as I have mentioned before, for the right-handed bivectors we would then have (B_i)(B_j)(B_k) = +1 , and for the left-handed bivectors we would have (B_i)(B_j)(B_k) = -1 . It is as simple as that. I see no inconsistency whatsoever of any kind, anywhere."

Nbody has posted the explicit caluclations so far. Although they are straightforward I'd like to do that: first I define a set of bivectors:

[equation]B_1^R=I\cdot e_1=e_1e_2e_3e_1=e_2e_3=e_2\wedge e_3\\

B_2^R=I\cdot e_1=e_1e_2e_3e_2=e_3e_1=e_3\wedge e_1\\

B_3^R=I\cdot e_1=e_1e_2e_3e_3=e_1e_2=e_1\wedge e_2\\[/equation]

for these bivectors the following relations hold:



and

[equation]B_1^RB_2^R=e_2e_3e_3e_1=-e_1e_2=-B_3\\

B_2^RB_3^R=e
_3e_1e_1e_2=-e_2e_3=-B_1\\

B_3^RB_1^R=e_1e_2e_2e_3=-e_3e_1=-B
_2\\[/equation]

which we can summarize as:



Furthermore we have the relation



According to Doran and Lasenby GA for Physicists p. 34 the last equations holds for a right-handed set of bivectors.

An example for a left-handed set of bivectors are quaternions, as pointed out in the same book at the same page.

[equation]B^L_1=i\\

B^L_2=j\\

B^L_3=k\\[/equation]

and obtain the products

[equation]B^L_1B^L_2=ij=k=B^L_3\\

B^L_2B^L_3=jk=i=B^L
_1\\

B^L_3B^L_1=ki=j=B^L_2\\[/equation]

and



the resulting algebra is therefore



And



which is called left-handed set of bivectors according to Doran and Lasenby.

Finally summarizing this results in a single equation gives:



it's exactly eq.4 in Joy's one-page paper

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John wrote on Sep. 9, 2011 @ 08:35 GMT
Dear all,

Joy mentioned in one of his replies to Florin:

"It is not all that mysterious to see if you really want to see it. Just separate out the two possible equations that are combined in equation (4) and see what they really are. One, with lambda = +1, is the even subalgebra of right-handed bivectors, and the other, with lambda = -1, is the even subalgebra of left-handed bivectors (originally proposed by Hamilton). And as I have mentioned before, for the right-handed bivectors we would then have (B_i)(B_j)(B_k) = +1 , and for the left-handed bivectors we would have (B_i)(B_j)(B_k) = -1 . It is as simple as that. I see no inconsistency whatsoever of any kind, anywhere."

Nobody has posted the explicit caluclations so far. Although they are straightforward I'd like to do that: first I define a set of bivectors:







for these bivectors the following relations hold:



and







which we can summarize as:



Furthermore we have the relation



According to Doran and Lasenby GA for Physicists p. 34 the last equation holds for a right-handed set of bivectors.

An example for a left-handed set of bivectors are quaternions, as pointed out in the same book at the same page.







We obtain the products







and



the resulting algebra is therefore



In addition



which is called left-handed set of bivectors according to Doran and Lasenby.

Finally summarizing this results in a single equation gives:



it's exactly eq.4 in Joy's one-page paper

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Joy Christian replied on Sep. 9, 2011 @ 08:57 GMT
John,

There are two minor errors in your first set of equations. But apart from that your calculations are fine. I am glad that at least someone paid attention to what I was saying.

Joy

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John replied on Sep. 9, 2011 @ 09:27 GMT
Joy,

I agree. My first set of equations should be:







John

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T H Ray replied on Sep. 9, 2011 @ 12:38 GMT
John,

Thanks! I hope others will join in to do the right thing.

Tom

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 17:42 GMT
John,

Excellent work! If only Joy would have done this himself and spare the acrimony. I will respond mathematically shortly (and not in 6 months on the achive), and we can continue to debate this the right way.

PS: your calculatons are all correct until combining it into a single equation. I will show why this is not allowed. There is a major difference between left and right algebras and left and right coordinate frames. Expect my reply within an hour.

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Lawrence B. Crowell replied on Sep. 9, 2011 @ 17:58 GMT
This is correct. John’s equations are also formally correct. First off the last two of the first set of equations he wrote clearly mean I*e_2 and I*e_3, but that does not impede things. In reading this I can't find any math error. However, the distinction between right and left handed bivectors defines the direction of multiplication. Quaterions with

ij = jk = ki = -1 = ijk

while left handed (left multiplication) ones are not. According to this definition the right handed bivectors are

B_1 = I*e_1 = e_2/\e_3,

B_2 = I*e_1 = e_3/\e_1,

B_3 = I*e_3 = e_1/\e_2,

while for left handed

B_1 = I*e_1 = ie_2/\e_3,

B_2 = I*e_1 = ie_3/\e_1,

B_3 = I*e_3 = ie_1/\e_2,

for i = aqrt{-1}. This then changes the sign of B_iB_j = (+/-)eps_{ijk}B_k. We have been arguing here with B_2 and -B_2 and so forth, but this change of signs is due to a factor of i = sqrt{-1} or a rotation e^{-i π/2} rotation e^{-i π}. This factor changes the meaning of multiplication for right to left multiplication is all. This is different from the notion of change in the orientation of a vector.

Cheers LC

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Ray Munroe replied on Sep. 9, 2011 @ 20:48 GMT
Dear Lawrence,

Thank you for the details. I think I have been saying similar stuff with different (more physical) terminology.

If I'm following you correctly, the right isoclinic rotations yield a quaternion 3-sphere with a space-like hypersurface (and a time-like radius), whereas the left isoclinic rotations yield a quaternion 3-sphere with a time-like hypersurface (and a space-like radius).

Improper mixing of this biquaternion thus mixes up space and time components, and may introduce complex numbers via square roots of the Minkowski metric (-1,+1,+1,+1). Complex wavefunction symmetries could then lead to Anyonic statistics, the faster-than-light speed of fundamental tachyons, and all sorts of problems for the question of Local Reality if these tachyons participate in local AND non-local interactions.

Dear John,

Thank you for the mathematical details.

Have Fun!

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Lawrence B. Crowell replied on Sep. 9, 2011 @ 23:10 GMT
The mulitplication B_iB_j will need to include an additional factor of i = sqrt{-1}. So for

B_1 = I*e_1 = ie_2/\e_3,

B_2 = I*e_1 = ie_3/\e_1,

The multiplication B_1B_2 must be

B_1B_2 = i(ie_2/\e_3)(ie_3/\e_1)

to be consistent with the above definitions of how B_i is formed with the rule ab = a*b + ia/\b for right multipication, here the i is included to make it quaternionic for a,b,c as quaternions with ab = a*b + a/\b. This is then

B_1B_2 = -ie_2/\e_1 = ie_1/\e_2 = B_3

This is a way of seeing how the left rule can be seen as an imaginary version of the right rule. This is the same as saying there is a right and left multiplication rule with

ab_R = ab = a*b + a/\b

ab_L = ba = a*b + b/\a = a*b – a/\b

There are also octonionic rules of multiplication as well, with the Jordon product

aob = (1/2)(ab + ba),

which leads to the Freudenthal product rule.

Florin illustrates how the mixing of the two results in troubles. The right bivector rule includes the elements (1, B_1, B_2, B_3) with

B_i^2 = -1, B_1B_2B_3 = 1,

If we treat the 1 in the elements as B_0 the metric has signature [1, -1, -1, -1]. However the 3-rule here does not conform to Hamilton’s rule

ijk = -1 = ij = jk = ki.

The left rule does.

Cheers LC

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Florin Moldoveanu wrote on Sep. 9, 2011 @ 20:59 GMT
Dear John,

Just a simple exercise proving I do have the same left and right algebras as you do.

Use:

B1R = e2e3

B2R = e3e1

B3R = e1e2

B1L = -e2e3

B2L = -e3e1

B3L = -e1e2

B1R B2R = e2e3 e3e1 = e2e1 = - e1e2 = -B3R (just like your case)

B1L B2L = (-e2e3) (-e3e1) = e2e1 = - e1e2 = +B3L (just like your case)

Please convince yourself that BiBj = -delta_ij - epsilon_ijk Bk for the right algebra

Please convince yourself that BiBj = -delta_ij + epsilon_ijk Bk for the left algebra

I wish I know how to use this mini LateX on this editor to present you the matrix representations of the left and right algebras and show that the mapping between them is the mapping between row and columns, or in QM notation between kets and bras. Mixing an matching left and right algebras is equivalent to adding apples and oranges, or kets and bras.

Handedness on the other hand (no pun intended) comes from a mirror symmetry. Geometric algebra is famous for easy representation of rotations, and in there a rotation is implemented by 2 consecutive mirror reflections against 2 distinct planes. If you need to switch between algebra representations to represent a mirror symmetry, then the algebra itself would not be powerful enough to capture rotations, and this is just not the case.

In summary, you have a left algebra with a left handed basis, a left algebra with a right handed basis, a right algebra with a left handed basis, and a right algebra with a right handed basis: 4 combinations. To be correct, all computations has to be done within one algebra.

HERE are the 4 combinations:

Here is the right (sub)algebra in a right handed basis:

{1, e2e3, e3e1, e1e2}

Here is the right (sub)algebra in a left handed basis:

{1, (-e2)e3, e3e1, e1(-e2)}

Here is the left (sub)algebra in a right handed basis:

{1, -e2e3, -e3e1, -e1e2}

Here is the left (sub)algebra in a left handed basis:

{1, e2e3, -e3e1, e1e2}

Proof:

right (sub)algebra in a right handed basis:

e2e3e3e1 = e2e1 = -(e1e2): B1B2 = -B3 => right algebra. Right handed by basis by drawing e1, e2, e3 on paper and applying right hand rule.

right (sub)algebra in a left handed basis:

(-e2e3)e3e1 = -e2e1 = (e1e2) = -e1(-e2): B1B2 = -B3 => right algebra. Left handed by basis by drawing e1, -e2, e3 on paper and applying left hand rule.

left (sub)algebra in a right handed basis:

(-e2e3)(-e3e1) = e2e1 = +(-e1e2): B1B2 = +B3 => left algebra. Right handed by basis by drawing e1, e2, e3 on paper and applying right hand rule.

left (sub)algebra in a left handed basis:

(e2e3)(-e3e1) = -e2e1 = +(e1e2) : B1B2 = +B3 => left algebra. Left handed by basis by drawing -e1, e2, -e3 on paper and applying left hand rule.

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Florin Moldoveanu replied on Sep. 9, 2011 @ 21:35 GMT
Oops, typo/mistake on last line above.

Instead of reading "drawing -e1, e2, -e3 on paper and applying left hand rule."

please read: "drawing e1, -e2, e3 on paper and applying left hand rule."

Why? because the transition from "left (sub)algebra in a right handed basis" to "left (sub)algebra in a left handed basis" is made by switching e2 to -e2 in the basis {1, -e2e3, -e3e1, -e1e2} -> {1, -(-e2)e3, -e3e1, -e1(-e2)} =

{1, e2e3, -e3e1, e1e2}

By the way, "left" in left-algebra means matrix multiplication to the left with a raw vector and "right" in right-algebra means matrix multiplication to the right with a column vector when representing the algebras in matrix format.

Left-handed and right-handed refers to the sense of rotation "moving" the first basis element of the basis unto the second element of the basis and determining if the rotation is positive or negative on the third basis vector.

Same left and rigth names, completely different meaning.

In multiple dimensions, there are still left and right algebras (bras and kets) but handedness applies only to 3D.

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James Putnam wrote on Sep. 9, 2011 @ 23:11 GMT
Dear Florin,

I am curious about your list of errors that you posted on the arxiv.org. Are they related to today's discussion or replaced by it? Are they still held to be errors?

James

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Florin Moldoveanu replied on Sep. 10, 2011 @ 00:11 GMT
Dear James,

I had posted 4 errors on my preprint. All the discussions so far with a minor exception were on error number 1. Error number 2 is about the impossibility of correcting error 1 without breaking another just as critical requirements for agreement with experimental results. Errors 3 and 4 are about additional mistakes by Joy in an analysis dedicated to why all four experimental outcomes (++, --, _+-,-+) are obtained by his model. Specifically, error 3 is about a wrong limit: zero/zero which Joy incorrectly claims is equal with zero, while error 4 is about an incorrect “rotor” geometric algebra object used to perform a rotation.

Today’s discussion is solely related to error number 1. But there are two aspects to it. In my preprint http://arxiv.org/PS_cache/arxiv/pdf/1109/1109.0535v1.pdf I pointed out how Joy made this error and I stated two things:

First:

“So how it was possible to have such an elementary mistake undetected? Most of Joy Christian’s papers suffer from a convention ambiguity: in some cases the computations are done using the μ = ±I convention with I arising from a fixed basis, while in others the computations are done using the μ = I convention”

This is related to today’s discussion of mixing left and right algebras.

Second:

“Another way this mistake can arise can be seen in Eqs. 23 and 24 of Ref. [6]. In there Eq. 23 is correct and Eq. 24 is derived by switching I to −I for a change of handedness. It is true that changing handedness (or equivalent performing a reflection) changes the sign of the pseudoscalar I, but what is incorrect in Eq. 24 is that a × b is a pseudovector who should change sign as well.”

This is related to the prior discussion about the bad sign of the Hodge duality in a left handed basis.

I am in the process of preparing a simple tutorial on the matrix representations for left and right algebras where the difference between the two is obvious.

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Florin Moldoveanu replied on Sep. 10, 2011 @ 00:27 GMT
Dear James,

By the way, I did not know how Joy made the mistake: incorrect Hodge duality, or mixing algebras and that is why I put both explanations in my paper. During the debate Joy first insisted on the Hodge duality and that is why I presented the math on that. When he lost that mathematical argument, he latched onto John’s computation and changed his tune: “I am glad that at least someone paid attention to what I was saying.” No, that was not what he was saying earlier, but instead of arguing with this, I am now trying to make the argument against swapping algebras as simple as possible. Either way, Joy’s error # 1 is debunked.

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Florin Moldoveanu wrote on Sep. 10, 2011 @ 03:11 GMT
Thear John,

This post is about left and right matrix representations of algebras. This is advanced linear algebra which cannot be found on the web or in any introductory college texts. Please request clarifications for any unclear parts.

Let's start simple: complex number algebra

Any complex number z can be represented as z = a+sqrt(-1) b

It can also be represented as...

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Florin Moldoveanu replied on Sep. 10, 2011 @ 03:14 GMT
Dear John,

Sorry about the typo, I must have been blind.

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Florin Moldoveanu replied on Sep. 10, 2011 @ 04:07 GMT
Dear John,

After re-reading my matrix representation post I realized I did not spell out the relationship with Clifford algebras. Let me do it now for complex numbers

Complex numbers in RIGHT matrix representation are:

Z_r = a + sqrt(-1) b = Matrix(a, b; -b, a)

Then define 1_r = Matrix(1, 0; 0, 1) – identity matrix

And define I_r = Matrix(0, 1; -1, 0) (sometimes called Omega)

Z_r = a 1_r + b I_r

Define e1 = Pauli_3 = Matrix(1,0;0,-1)

Define e2 = Pauli_1 = Matrix(0,1;1,0)

Then

e1e2 = I_r

e2e1 = -I_r

e1e1 = e2e2 = 1_r

And the RIGHT subalgebra is {1, e1e2}

~~~~

Complex numbers in LEFT matrix representation are:

Z_l = a + sqrt(-1) b = Matrix(a, -b; b, a)

Then define 1_l = Matrix(1, 0; 0, 1) – identity matrix

And define I_l = Matrix(0, -1; +1, 0) (sometimes called -Omega)

Z_l = a 1_l + b I_l

Define e1 = Pauli_3 = Matrix(1,0;0,-1) –same as above

Define e2 = Pauli_1 = Matrix(0,1;1,0) – same as above

Then

e1e2 = -I_l

e2e1 = +I_l

e1e1 = e2e2 = 1_l

And the LEFT subalgebra is {1, -e1e2}

Also

1_r = 1_l

I_r = - I_l

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Florin Moldoveanu replied on Sep. 10, 2011 @ 11:48 GMT
Dear John,

A quick note. I have an inconsistent definition on why the matrix representations of the algebras are called left and right. The latest definition is correct and in line with literature, the former it is not (as I was quoting it from my head and I did not double check in any book at the time of writing because I did not have any books available). Sorry about this confusion. To recap, I called a representation incorrectly RIGHT for multiplying a right Ket, when in fact in literature it is called LEFT for multiplying the Ket on the left. Same thing with the Bra.

The QM analogy is not an accident, but runs very deep and the commutation rule between left and right representations allows Leibniz rule of derivation: D(A B) = D(A) B + A D(B) from which all calculus follows. Therefore one can consider the RIGHT representation of an associative algebra to represent derivations of the LEFT representation of an associative algebra (or the other way around).This is equivalent with the momentum and position representation in QM. In general Left and Right algebras combined are not algebraically stable and the lack of stability when derivations are in the RIGHT representation and the functions are in the LEFT representation has a deep relationship with gauge theory and the Berry geometric phase.

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Florin Moldoveanu wrote on Sep. 10, 2011 @ 15:14 GMT
This post presents the 4 possible combinations: Left and right algebras with left and right handedness and hopefully dispels the confusion between left/right dichotomy in algebras and handedness and clearly illustrate mathematical error #1 in Joy’s papers.

Consider {e1,e2, e3} a right-handed base: let e1 point towards you away from the paper, e2 point to the right and e3 point...

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Florin Moldoveanu replied on Sep. 10, 2011 @ 20:29 GMT
All the 4 examples above are realizations of the same Clifford (sub)algebra product:

Bi Bj + Bj Bi = - 2 delta_ij

just like the left and right representations of complex or quaternion algebras are realizations of the same multiplication rule.

(the original Clifford algebra product has a + sign in front of the delta, but here B are bivectors of negative norms as oppose to vectors of positive norms)

It is rather convoluted to compute the 4x4 matrix representation of those 4 realizations (either over reals or over complex numbers) but it can be show easily that the name indeed does come from left and right matrix representation by comparison with complex numbers which require a much simpler 2x2 representation.

Right complex number algebra:

z = a+sqrt(-1) b = {1*a, e1e2*b} = Matrix(a,b;-b,a)=Row(a,b)

In Clifford algebra terms: {1, e1e2}

Left complex number algebra:

z = a+sqrt(-1) b = {1*a, -e1e2*b} = Matrix(a,-b;b,a)=Column(a,b)

In Clifford algebra terms: {1, -e1e2}

All with

e1 = Pauli_3 = Matrix(1,0;0,-1)

e2 = Pauli_1 = Matrix(0,1;1,0)

Compare

{1, e1e2}-Right algebra

{1, -e1e2}-Left algebra

with:

{1, B1=e2e3, B2 = e3e1, B3 = e1e2}-Right algebra (right handedness)

{1, B1=-e2e3, B2 = -e3e1, B3 = -e1e2}-Left algebra (right handedness)



Switching between algebras, or mixing them in the middle of computation is illegal and results in nonsensical answers.

Joy claimed earlier (Sep. 5, 2011 @ 13:16 GMT) “My equation (4) is not incorrect. It is a postulate, not a deduction from equation (3). A postulate cannot be incorrect.” While referring to Eqs 3 and 4 of http://arxiv.org/PS_cache/arxiv/pdf/1103/1103.1879v1.pdf

One cannot postulate 2 different algebras acting on the same domain at the same time because you get either (A) inconsistent results, or (B) another algebra if you get lucky. This is not case (B) because assuming both algebras (3) and (4) at the same time, the left hand side of both Eq. 3 and Eq. 4 are identical meaning their right hand side should be the same as well. What does this mean?

-delta_jk –epsilon_jkl Beta_l = -delta_jk – lamda*epsilon_jkl Beta_l

simplify –delta_jkabd bring all terms on the left hand side:

-epsilon_jkl Beta_l *(1-lambda) = 0

epsilon_jkl Beta_l is never zero and the only way this is true is if:

lambda = 1

However, Joy also postulates “lambda = +/-1” and we arrive again at the contradiction +1 = -1

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Florin Moldoveanu replied on Sep. 11, 2011 @ 04:07 GMT
One more thing to justify the names LEFT and RIGHT algebras for complex numbers

Let z be:

z= a + i b, i = sqrt(-1)

The LEFT algebra for complex numbers in matrix and ket form is:

Z = Matrix(a, -b; b a)

|z) = Column(a, b) – a Ket vector

Z |w) = |z *w) with * the standard complex multiplication

Proof:

z = a + i b

w=c + i d

z*w = ac-bd + i (ad+bc)

|z*w) = Column (ac-bd; ad+bc)

Z |w) = Matrix(a, -b; b,a) Colum(c;d) = Column(ac-bd; bc+ad) – same as above

Now this can be generalized to:

A B C |d) = A B |c*d) = A |b*c*d) = |a*b*c*d) – multiplication from the LEFT on kets or column vectors

~~~~

The RIGHT algebra for complex numbers in matrix and bra form is:

Z = Matrix(a, b; -b a)

(z| = Row(a, b) – a Bra vector

(w| Z = (w*z | with * the standard complex multiplication

Proof:

z = a + i b

w=c + i d

z*w = ac-bd + i (ad+bc)

(z*w| = Row (ac-bd, ad+bc)

(w| Z = Row (c, d) Matrix(a, b; -b,a) = Row (ca-db, cb+da) – same as above

Now this can be generalized to:

(a| B C D = (a*b| C D = (a*b*c| D = (a*b*c*d| – multiplication from the RIGHT on bras or row vectors

There is also a 1-to-1 mapping between Kets, Bras, Left and right matrix representation for ALL associative algebras.

Because matrix multiplication is associative, all associative algebras can be put in matrix form. The associative rule is equivalent with the commutator between left and right matrix representations: [A_left, A_right] = 0

Kets and bras are mapped one to another by the Hermitean conjugate. Matrix representations exist over real or complex numbers.

Mixing algebra representations is equivalent to mixing kets and bras, column and row vectors – a complete nonsense which is not as clear in other representations like the standard representation of Clifford algebras.

Only in 3D Clifford algebras have an additional property: handedness which is a property of the cross product and its Hodge duality with bivectors.

Left and right algebras are also linked by vector space duality as one can define functions in one, and derivation operators in the other. When applying the say left derivation onto a right algebra of functions, the lack o closure gives rise to gauge symmetry and geometric phases (Berry phase), in other words, there is a natural way to introduce curvature.

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Eckard Blumschein replied on Sep. 11, 2011 @ 11:14 GMT
Dear Florin,

Please do no reject my arguments without careful scrutiny: I consider it possible and in case of dealing with reality necessary to restrict to only measurable i.e. elapsed time. Without negative elapsed time there is no reason to use complex

Fourier transformation and Hermitian symmetry instead of cosine transformation and half matrices.

This implies that i is only required for Heaviside's detour. While there is no non-commuting conjugate variables, there are nonetheless the well known features of conjugacy between e.g. time and frequency, see my essays.

If Joy is wrong, and I also think so since he postulates time-symmetry, then the reason might be in a mistake of QM.

Regards,

Eckard

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Florin Moldoveanu wrote on Sep. 11, 2011 @ 14:23 GMT
Dear Eckard,

Let me start another thread to answer your question. If I understood you correctly, you are arguing for a real formalism (not using complex numbers). Indeed, for Fourier analysis one can work equally well with sin(alpha) and cos(alpha) instead of exp(i alpha), it is all a matter of convenience. For example, doing basic circuit theory analysis for resistors, capacitors,...

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Eckard Blumschein replied on Sep. 11, 2011 @ 20:44 GMT
Dear Florin,

Thank you for dealing with what I am considering overlooked. Admittedly, I do not yet understand spin and its role in QM. Was the experiment by Stern and Gerlach the decisive basis for Heisenberg, Schroedinger, and Dirac? I will look into the sources. If I recall correctly, spin was at least initially imagined in connection with hypothetic moving electrons which were thought to belong to a magnetic moment.

120°+120°+120°=360°=0°

60°+60°+60°=180°= negative direction

I already checked a book that was recommended to me by Eugene Klingman: An Imaginary Tale by Nahin.

Perhaps you did not yet get the essence of my reasoning. I do not split a complex representation into sine and cosine part. I see the cosine transformation of a function of elapsed time not just a particular case of Fourier transformation but fully adequate. Let me tell you why:

I refer to the elapsed time in each moment measured relative to this moment, as we do when speaking of the currently increasing age of something. Admittedly, such sliding, relative to the events perspective, scale is quite unusual. However, its natural zero of just elapsed time corresponds to the natural zero of radius. The otherwise unavoidable arbitrariness of choosing the point t=0 disappears. Consequently, the sin component equals to zero. Surprisingly such permanent referring to the point between past and future makes the complex representation indeed redundant.

From this perspective, the complex approach is an unnecessary in principle while sometimes superior detour. This also includes non-commuting variables and Hermitian symmetry as requisites of the complex detour. Apparent time-symmetry is a quite understandable artifact that was misinterpreted by those who did not understand what they were doing.

Regards,

Eckard

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Florin Moldoveanu replied on Sep. 12, 2011 @ 00:46 GMT
Dear Eckard,

You say: “I see the cosine transformation of a function of elapsed time not just a particular case of Fourier transformation but fully adequate.”

Fully adequate to what? Fourier modes are functions which form a vector space where any arbitrary function can be decomposed into. There are many such other series of functions: Bessel, spherical harmonics, etc. Also those functions have to satisfy 2 basic properties: (a) each function has to be orthogonal on all others in the series, (b) completeness: any arbitrary function should be written as a (infinite) sum of the function in the series.

Now only the cosines are not a complete basis because there are functions which cannot be written as a sum of only cosines. A trivial example is any sine function. Why? Sin(x) = -Sin(-x) and Cos(x) = +Cos(-x) and a anti-symmetrical function like sine cannot be written as a sum of only symmetric functions.

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Eckard Blumschein replied on Sep. 12, 2011 @ 04:38 GMT
Dear Florin,

I maintain: Cosine transformation of any continuous function of x larger than zero and smaller than infinity is fully adequate to Fourier transformation.

Even functions that come as close as you like to the ideal sin function can be decomposed into an infinite series of cosine components with appropriately chosen positive and negative amplitudes. Admittedly, these components have to be larger and larger the closer one tries to approach the ideal sinus.

This particularity is known to cause poor convergence in Fourier Acoustics, cf. the textbook by Earl Williams. It can be understood by every cyclist. There is a dead point exactly at angle zero to the vertical - no obstacle for cycling. Buridan's ass and mathematicians are less practical. Their inability to unequivocally deal with the very point zero between R+ and R- was an enigma to me until I found out that they left the path of Euclidean virtue. I already earlier understood that a single point in R has no measure. Nobody would be able to decide whether or not it was stolen if this would be possible at all. Such considerations are not even required because elapsed time steadily changes.

Regards,

Eckard

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Ray Munroe wrote on Sep. 13, 2011 @ 00:13 GMT
Hi Tom,

Per Florin's suggestion, I thought we should disentangle our conversation from Florin's and Anonymouse's.

Please correct me if I'm wrong, but doesn't Joy supposedly work with an S^7 7-sphere? I figure that the following GA signatures may have legitimate physical analogies with the 7-sphere:

G(8,0) - A real 8-ball,

G(7,1) or G(1,7) - an octonion 7-sphere,

G(6,2) or G(2,6) - a bi-quaternion, or

G(4,4) - a complex bi-quaternion.

The only option that does not introduce complex numbers is the G(8,0) 8-ball - which is an 8-D extension of 3-D Euclidean geometry, but nothing here reflects an obvious 7-fold symmetry, or anything resembling the properties of time.

The only option that has a clear 7-fold symmetry and a single time coordinate is the octonion 7-sphere.

The bi-quaternions could be physical, but they all contain 2 or more time dimensions, and do not contain obvious 7-fold symmetries.

Which GA signature are you and Joy using? Choose your poison - I think I can tear down most of these options. Once we understand which 7-sphere we are using, then we can try to deduce the interface between this 7-sphere and 3-D Euclidean space.

Have Fun!

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Ray Munroe replied on Sep. 13, 2011 @ 12:02 GMT
No answer...

Are the proponents of Joy's theory afraid to step into a position that they know they can't defend?

Have Fun!

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T H Ray replied on Sep. 13, 2011 @ 12:33 GMT
Hi Ray,

Getting you to think classically has become quite a challenge. :-)

First thing, note that Hestenes's spacetime algebra (STA) which follows from geometric algebra (GA) which follows from vector algebra which follows from complex analysis -- is written such that the formalisms can be translated into the Minkowski spacetime. This is not possible in either complex analysis or vector algebra. So when we convert the analytical language of continuous functions to the algebraic language of discrete functions, we have necessarily built in a time dependency of dynamic interaction. That's how Joy derives a fixed bivctor basis and a fluctuating bivector basis, which means that while all GA results are necessarily real, an indeterminate measure is the hidden variable; i.e., instead of assigning value to a probabilistic measure ("the experiment not performed"), the analytical case of discrete measure assigns the value "undetermined" and integrates that value into a fully deterministic continuous function model.

I admit (it's on record at the start of the discussion here) that I also reacted to the mention of S^7 much the same way as you are. I expected that the model had to be nonphysical. After some study, I realized that the only signifance of S^7 was to orient the measure to points of S^3. In the end, all that mattered for a continuous function analysis was the choice of topology and the initial condition that prescribed all the information up to and including the cosmoloigcal condition.

Joy was very clear about the orientation of his results on the equator of S^3. Knowing that the only results that live there are + 1, - 1 and i, I showed (slide 11 in my proposed experiment pdf under the heading "completeness") that vector rotations are all real and 3 dimensional. Between i and i' we find:

(+ 1, - 1, i )^1 = i

(+ 1, - 1, i )^2 = {+ 1, + 1, - 1}

(+ 1, - 1, i )^3 = {+ 1, - 1, + 1}

(+ 1, - 1, i )^4 = {+ 1, + 1, - 1}

(+ 1, - 1, i )^5 = i'

As Hestenes says in the spacetime algebra paper: "... we provide the real Dirac wave function with a geometric interpretation by relating it to local observables. The term “local observable” is non-standard but the concept is not unprecedented. It refers to assignment of physical interpretation to some local quantity such as energy or charge density rather than to global quantities such as expectation values." ("VII. Real relativistic quantum theory," p. 31)

Tom

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Ray Munroe replied on Sep. 13, 2011 @ 14:08 GMT
Hi Tom,

I understand that Bell's derivation was 'classical' and 'non-relativistic', but he didn't appeal to a 7-sphere that - we might at least expect - should contain Spacetime as a subset. The G(8,0) 8-ball does not contain spacetime as a subset. But all other possible GA signatures contain complex numbers and spacetime as a subset. Therefore the reasonable conclusion is that the 7-sphere introduces complex analysis.

I think that we are saying the same (or at least similar) thing with regard to the fact that spacetime looks 'real' when we embed it in the Minkowski metric - and we both recognize that the Minkowski metric contains quaternion physics. This is the basis of my claim that if coordinates are mixed in an incorrect manner (and I'm not clear if Joy is setting 'fixed time' between different GA signatues - apples and oranges, or if Joy is discarding time because Bell discarded time in a more simple model), then we introduce the need for complex numbers in our analysis - it is no longer strictly real. In fact, you have an 'i' that you discard based on your personal conclusion "that vector rotations are all real and 3 dimensional".

If we introduce complex numbers into Joy's analysis, I would expect this to increase all mean values by the SQRT(2) based on a degrees of freedom argument. Please recall that BELL'S THEOREM is a THEOREM! Whether I personally like the idea of local reality or not is irrelevant - any mathematician wanting to overthrow a THEOREM needs to 'dot his i's' and 'cross his t's' - no sloppiness allowed!

Have Fun!

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Steve Dufourny wrote on Sep. 14, 2011 @ 17:02 GMT
ahahaha the poor thinkers, ahaha

and after what invite me, drink with me, eat with me and travel with me and after you shall kill me , a band of business men simply. Without faith and consciousness ,

but in fact and fortunally this universal sphere is a project for good people , they shall fall down simply here on earth or after it is not a problem for this universal sphere you know.

To be or not to be indeed.

Even dead I will continue my works , yes i am crazzy .

Incredible example of the human nature and its taste for monney, power, vanity and guns.

Steve

Steve parano but consciouss of his discovery.

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Steve Dufourny the sphere replied on Sep. 14, 2011 @ 21:24 GMT
you want really a crazzy ,parano who says the truths....i do not fear because i have faith .

Steve

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Steve Dufourny wrote on Sep. 14, 2011 @ 17:08 GMT
here is the team, not complete of course but it is a begining of course.

Lawrence B Crowel, Gareth Lisi,TH Ray, Ray Munroe,Joy Christian, Florin Moldeveanu,.....still 2 and 8 others behind also perhaps .And a E16 after ahahah beautiful strategy.

Study my new equations !!!

Steve

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Edwin Eugene Klingman wrote on Sep. 15, 2011 @ 00:53 GMT
Dear All,

This blog represents FQXi at its finest. Brilliant minds discussing foundational questions.

In an earlier blog, Florin said, "Indeed, Joy did not disprove Bell's theorem as this theorem remains mathematically correct" while (at that time) claiming Joy's work also mathematically correct.

Having studied a hundred page printout of the above comments,I do not yet know whether Joy's math is correct, as very competent people disagree over this. I quote some very interesting comments:

TH Ray: "so we go from analysis to algebra to geometric algebra in a smooth way."

Steve Dufourney: "The aim is not to add several mathematical methods in a kind of architecture, but to use these tools for rationally to interpret the physicality!"

Ray Munroe: "The mathematical logic *MIGHT ALMOST* work...but the physical logic is clearly defective."

Rob Onestone: "Mathematics should be correct, no matter what physical interpretation you gave to it."

Joy: "After all, the model is supposed to mimic what happens in the actual physical experiment."

Florin: "We can put the math arguments on hold and discuss the physical arguments instead."

Finally, several of you--"I'm tiring of this argument."

Assume that both are mathematically correct. (I think this is Tom's conclusion.) What does this mean? It tells me that (at least) one is physically incorrect. This approach is for the most part missing from the above arguments. I have written a few pages on this that are relevant. Because of equations and diagrams, I have not attempted to fit them into a comment but placed them here.

I would very much appreciate comments on this approach.

Edwin Eugene Klingman

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Jason Wolfe replied on Sep. 15, 2011 @ 01:35 GMT
Hi Edwin,

Just to comment on quantum entanglement, I picture it as a "fiber made out of space-time" in such a way that photons (information) will travel back and forth at the speed of light relative to each of the two particles. Originally, I had an image of chewing gum connecting my foot to the ground.

The quantum mechanics doesn't need to place a velocity limit on the quantum entanglement. The invariance of the speed of light already does that for us.

Quantum entanglement, if it wasn't so fragile, is like a hybrid of wireless communication and copper line. The photons should use the connection like a freeway, back and forth at the speed of light, until the connection breaks.

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T H Ray replied on Sep. 15, 2011 @ 08:49 GMT
Edwin,

You conclusion does not follow. For about 2000 years, Euclidean geometry was considered synonymous with the physical world. It still works for most practical applications, and it is completely mathematically correct in its domain. Newtonian physics is mathematically correct and entirely adequate for even highly technological purposes, such as landing a person on the moon.

Non Euclidean geometry and relativistic physics are mathematically complete. That is, they are able to make closed judgments on the nature of reality independent of experiment. Quantum mechanics is not mathematically complete in this way (though quantum field theory seeks to be); it is founded on explaining phenomena a posteriori. Clearly, this is why Florin has taken the fall-back position of trying to fit Joy's model into a Procrustean bed of probabilistic experimental results. This goes against the very purpose of Joy's model, a mathematically complete judgment that includes the predictions of QM and trivializes Bell's theorem. Not even Joy says there's anything wrong with Bell's mathematics, just his initial assumption, which, while it validates the theorem, makes it the wrong choice on which to base a complete model of physically reality.

Tom

Tom

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T H Ray replied on Sep. 15, 2011 @ 13:35 GMT
Edwin,

Apologies. I didn't see your link and I just read your draft this morning. Nice job, and good luck with it. I just want to point out that both Joy's and Steinberg's models -- while they do promote 1 to 1 correspondence between elements of theory and elements of experiment -- don't quite match your binary true/false criteria. Theirs is a 3-valued logic: true, false and undetermined.

Tom

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Florin Moldoveanu wrote on Sep. 15, 2011 @ 11:48 GMT
Tom,

In order to discuss the merits or validity of a physical theory one needs to be able to use it and generate experimental predictions. Those predictions are then compared against experimental results. Otherwise we will be still discussing how many angels can dance on the head of a pin.

Now please use Joy’s model any way you see fit and please predict the experimental outcomes of a spin measurement on the z direction followed by a measurement on the z or the x direction. Experimentally nature has the following results: zz: the outcome remains the same on the second measurement, zx: the outcome on x is 50% up 50% down. Does Joy’s theory predict the same thing or not. If not, it is a model which is not in agreement with nature and should be discarded. If yes, please show how Joy’s model agrees with experiments.

By the way, this is not my argument, this was originated by Holman, and I only made a simplification on the argument.

So Tom, let’s stop discussing how many angels can dance on the head of a pin and have physical a good old fashion physical argument instead. If you cannot use Joy’s model to predict experimental outcomes (assuming his math is correct-which is not, but let’s do it for the sake of the argument), then you are not qualified to issue judgments on its validity.

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T H Ray replied on Sep. 15, 2011 @ 13:07 GMT
Florin,

In fact, I did propose an experiment by Joy Christian's criteria. I'll attach it again, so you don't miss it.

This experimental model shows that Bell-Aspect predictions are recovered on the fixed axis, so it does predict the same thing. THAT'S THE POINT -- it HAS to. Which makes your argument completely superficial and not at all in accordance with a continuous function model of the real world.

The completeness of the Joy Christian framework allows further predictions of an objective initial condition, with random variables off the axis, which differs from the Bell-Aspect observer-chosen initial condition with probabilistic (nonlocal) outcomes. If Joy is right, these predictions are inetgrable, of Lebesgue (i.e., real analytical) measure, and completely deterministic and local even though over an indefinite spacetime interval (you never have understood Joy's use of the term "indefinite" in the right context) -- because the initial condition holds right up to the limit of the cosmological. So the integral from - oo to + oo is valid on any arbitrary interval of time, which obviates the notion of completeness attached to discrete "at a time" measures in which the interval [0,1] is fixed for all time and reality is observer-created. Joy's framework allows for the objective physical reality of spacetime, totally consistent with general relativity, while preserving the algebraic methods which define discrete measure without losing the generality of analysis, by using the language of geometric algebra.

You can think of it as angels and pinheads if you wish. Macht nichts to me.

Tom

attachments: Local_Realism_Experiment.pdf

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Florin Moldoveanu replied on Sep. 15, 2011 @ 13:29 GMT
Tom,

You are not answering directly to the question/challenge. I am asking what is 1+1 and you are answering cos*cos+sin*sin = 1. The question I am asking was selected precisely because Joy’s model predicts things in contradiction with experiment and has to be answered directly. It is not hard, and I can present the solution later. But my experience tells me that people who look knowledgeable from talking big things have terrible problems answering elementary questions. For a person who discussed all those complicated Hestenes’ ideas, it should be a piece of cake to extract experimental predictions from Joy’s theory. Unless you do not know how to do it and in this case just admit you cannot say that Joy’s theory is valid or not. Time to show your cards, I am calling your bluff.

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T H Ray replied on Sep. 15, 2011 @ 13:40 GMT
Florin,

You just don't get it. Joy is not predicting things in conflict with experiment. He is building in variables that explain the outcome of the Bell-Aspect coin toss predictions in a