Gordon Watson wrote:.
The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."
Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.
No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.
Reason:
(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).
(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).
To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).
I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.
Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216
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gill1109 wrote:Gordon Watson wrote:.
The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."
Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.
No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.
Reason:
(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).
(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).
To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).
I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.
Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216
.
Your logic is faulty, Gordon. I found your paper extremely difficult to read. You succeed in making simple things very difficult, and this enables you to make incorrect logical deductions.
It is indeed all extremely elementary. No knowledge of QM is needed to follow Bell's arguments, though of course sooner or later, one should verify the classical QM derivation of the EPR-B correlations, by matrix algebra calculations (over the complex numbers) involving the 2x2 Pauli matrices and various 4x4 matrices built up from them. Clifford algebra and Geometric Algebra are sophisticated and beautiful mathematical structures which one can use to elegantly reproduce classical QM calculations. Joy Christian uses them for other purposes.
I don't understand what your problem is with "Richard's probability theory". Bell does an elementary probability calculation using the notations and terminology of 60's physicists since he was a physicist writing in the 60's for physicists.
But to get to the heart of the matter, you compare a new inequality of your own, labelled (5) in your new paper, to Bell's original three correlations inequality, reproduced and labelled (6) in your new paper. You note that a particular set of three correlations E(a, b) = -1/2, E(b, c) = -1/2, E(a, c) = +1/2 violates Bell's inequality but satisfies yours.
The reason they violate Bell's inequality is actually because it is impossible to find functions A(a, lambda), B(b, lambda), and rho(lambda) satisfying the usual constraints *and* such that E(a, b) = int A(a, lambda)B(b, lambda) rho(lambda) d lambda, etc. If you had wanted to disprove Bell by a counterexample featuring these values of the three correlations, you would have had to exhibit functions A, B and rho which reproduce your three correlations E while at the same time satisfying the standard requirements (A and B take values +/-1; rho is nonnegative and integrates to 1). But you don't do this! So your paper is empty of any content! It just exhibits muddled thinking!
I suggest you do try to find a concrete example of functions A, B and rho which do the job you want them to do. (Joy Christian thinks he did manage to do that).
I can tell you in advance that you will not succeed, since Bell's inequality is a true and trivial result in probability theory, known already by George Boole in the 1850s, and since then independently discovered again and again by many others, as Karl Hess points out (yet again) in his most recent paper.
If you do succeed, then please program the resulting model and let's do my computer challenge. 10 thousand Euro's says you'll fail.
Gordon Watson wrote:However, didn't Bell make the claim (or imply) that such functions did not affect his result? That he need only work with the expectations to prove his point?
Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.
Gordon Watson wrote:As for your computer challenge: Can I first program the two computers to move toward each other under the influence of their masses? Might you see such a challenge (for starters) as equivalent to your Bell-based challenge? And worth the prize-money?
gill1109 wrote:Gordon Watson wrote:However, didn't Bell make the claim (or imply) that such functions did not affect his result? That he need only work with the expectations to prove his point?
No, he did not make any such claim. His logic is very transparent, I think. The question is whether or not such functions could exist. Assume they do exist, and assume QM predictions are at least approximately true, leads to a contradiction. Therefore they do not exist.
gill1109 wrote:Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.
Sure, I hope to come back to that.
Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.
FrediFizzx wrote:Yeah, I don't know why Gordon keeps harping about some error in the derivation of the inequality. There is no error in the inequality itself as it is mathematically proven. The error arises when you say a, b and c can happen all at the same time. They can't. It is that simple.
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Heinera wrote:There is of course nothing wrong with the equivalence between (14a) and (14b).
Perhaps it will be clearer to Gordon if we take it step by step in the opposite direction, from (14b) to (14a):
We start with (14b):
Simply multiply out to get
Now since must equal 1 we simplify the above to
.
Swap the two product terms:
and put the minus outside the expression:
Bell Interview in Omni Magazine wrote:Bell: Then in 1932 [mathematician] John von Neumann gave a “rigorous” mathematical proof stating that you couldn’t find a nonstatistical theory that would give the same predictions as quantum mechanics. That von Neumann proof in itself is one that must someday be the subject of a Ph.D. thesis for a history student. Its reception was quite remarkable. The literature is full of respectable references to “the brilliant proof of von Neumann;” but I do not believe it could have been read at that time by more than two or three people.
Omni: Why is that?
Bell: The physicists didn’t want to be bothered with the idea that maybe quantum theory is only provisional. A horn of plenty had been spilled before them, and every physicist could find something to apply quantum mechanics to. They were pleased to think that this great mathematician had shown it was so. Yet the Von Neumann proof, if you actually come to grips with it, falls apart in your hands! There is nothing to it. It’s not just flawed, it’s silly. If you look at the assumptions it made, it does not hold up for a moment. It’s the work of a mathematician, and he makes assumptions that have a mathematical symmetry to them. When you translate them into terms of physical disposition, they’re nonsense. You may quote me on that: the proof of von Neumann is not merely false but foolish.
minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).
Heinera wrote:minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).
You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.
Joy Christian wrote:Heinera wrote:minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).
You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.
If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.
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Heinera wrote:Joy Christian wrote:Heinera wrote:minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).
You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.
If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.
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As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.
Joy Christian wrote:Heinera wrote:Joy Christian wrote:If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.
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As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.
No, you wrote that "A is not a random variable." You wrote something that is obviously wrong. Admit it.
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Heinera wrote:Joy Christian wrote:Heinera wrote:Joy Christian wrote:If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.
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As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.
No, you wrote that "A is not a random variable." You wrote something that is obviously wrong. Admit it.
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You need to learn to distinguish between A and A(a, lambda).
Joy Christian wrote:Heinera wrote:You need to learn to distinguish between A and A(a, lambda).
That is just nonsense. You are trying to justify your earlier nonsense with more nonsense.
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Heinera wrote:Joy Christian wrote:Heinera wrote:You need to learn to distinguish between A and A(a, lambda).
That is just nonsense. You are trying to justify your earlier nonsense with more nonsense.
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No, it is not nonsense. A is a function, without any reference to a random variable. On the other hand, A(a, lambda) is a function of a random variable, assuming lambda is a random variable.
But you have anyway completely missed the point here; it is the omission of lambda in minkwe's argument that makes it misguided.
Joy Christian wrote:That is nonsense. And it speaks volumes. It is you who is missing minkwe's point, and then trying you justify your faulty understanding with more and more nonsense.
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