"Bell's theorem refuted" now published by EPL

[Joy Christian]

Your model is manifestly nonlocal in Bell's sense.

Can you show explicitely where my model is nonlocal in Bell's sense?

No. It is not my obligation to show you where your model is nonlocal in Bell's sense. It is your obligation to write down the functions A(a, h) and B(b, h) as defined by Bell, in the universally accepted notation, without any handwaving verbal arguments that you usually put forward. Until you can do that, no one can, will, or should believe that your model is local in Bell's sense.
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[Austin Fearnley]

I think that the model is local, but it is not clear how measurement B is made. To see it as local, take equations 4 and 4a as defining how to measure A. For the B measurement, one replaces δ (= α - ϕ) by δ' (= β - ϕ) so as to reference the polarisation setting of Bob's detector when defining the B measurement. Next treat all estimations after equation 4a as really a post-experiment explanation of why the Bell correlation is [or is not] achieved. I suspect that is what they really are and, at that stage, alpha and beta can be used together.

This could be a local model checkable using a computer simulation. The model uses the formula cos^2 which is a nod towards Malus's Law, but IMO a computer simulation would fail to give the Bell QM correlation. The best test is to calculate A and B measurements and correlate them for a sample of particle pairs.

[Esail]

It is not my obligation to show you where your model is nonlocal in Bell's sense. It is your obligation to write down the functions A(a, h) and B(b, h) as defined by Bell, in the universally accepted notation.
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Bell's locality condition that functions A(a, h) and B(b, h) have to exist is not necessary for locality. This not to see is Bell's mistake.
My model with A(delta, h) and B(delta, h) meets Einstein's locality condition as shown above and repeated below. As we are talking about physics and not about mathematics anybody who claims to disprove my model has to prove that it does not meet Einsteins locality condition.

Local means that the measurement results do not depend on superluminal communication between the two sides. (Einstein locality condition)
So imagine Bob with his polarizer PB is on the earth and Alice with her polarizer PA is on the moon, 1 light second apart from the earth. Bob measures a photon 2 having the parameter lambda0. He then calculates a list of possible outcomes at A for any possible setting angle α of Alice’s polarizer. He then stores this list in a safe 0.1 second after he obtained the result. 0.5 seconds after the peer photon 1 has left the earth Alice sets her polarizer to α0 and sends this information to Bob via a classical channel.
After Bob gets this information he opens the safe and picks the outcome from the list for the submitted value of α0. A match occurs if this outcome is +1.

So the outcome for A is determined (at the time of storing the list in Bob’s safe) even before Alice has obtained her measurement results. And the outcome for B is defined before Alice has set her polarizer.

This is clearly local as the measurement results do not depend on superluminal communication between the two sides!

[Joy Christian]

It is not my obligation to show you where your model is nonlocal in Bell's sense. It is your obligation to write down the functions A(a, h) and B(b, h) as defined by Bell, in the universally accepted notation.

Bell's locality condition that functions A(a, h) and B(b, h) have to exist is not necessary for locality. This not to see is Bell's mistake.

That is complete nonsense. You have no clue what you are talking about. I suggest you first educate yourself about the problem Bell correctly recognized before making such absurd claims.
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[Esail]

Bell's locality condition that functions A(a, h) and B(b, h) have to exist is not necessary for locality. This not to see is Bell's mistake.

That is complete nonsense. You have no clue what you are talking about. I suggest you first educate yourself about the problem Bell correctly recognized before making such absurd claims.
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That is not nonsense as my model correctly predicts the QM correlations. Also it has been proved that it is local in Einsteins sense.
If you still mean otherwise you should back it up instead of making unsubstantiated claims.

[Joy Christian]

Bell's locality condition that functions A(a, h) and B(b, h) have to exist is not necessary for locality. This not to see is Bell's mistake.

That is complete nonsense. You have no clue what you are talking about. I suggest you first educate yourself about the problem Bell correctly recognized before making such absurd claims.

That is not nonsense as my model correctly predicts the QM correlations. Also it has been proved that it is local in Einsteins sense.
If you still mean otherwise you should back it up instead of making unsubstantiated claims.

I do not have to back up anything. You are the one who is making a fool of yourself and everyone but you can see that. No one is surprised by a nonlocal model like yours reproducing the QM correlations. Bell himself showed that in an explicit model in his 1964 paper that a nonlocal model like yours can reproduce QM correlations. Einstein locality is not the issue here. You claim what you are claiming because you have no idea what the real problem is. I again suggest -- for your own sake -- that you educate yourself about what the real problem Bell had recognized.
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[Heinera]

[...]
So imagine Bob with his polarizer PB is on the earth and Alice with her polarizer PA is on the moon, 1 light second apart from the earth. Bob measures a photon 2 having the parameter lambda0. He then calculates a list of possible outcomes at A for any possible setting angle α of Alice’s polarizer. He then stores this list in a safe 0.1 second after he obtained the result. 0.5 seconds after the peer photon 1 has left the earth Alice sets her polarizer to α0 and sends this information to Bob via a classical channel.
After Bob gets this information he opens the safe and picks the outcome from the list for the submitted value of α0. A match occurs if this outcome is +1.

So the outcome for A is determined (at the time of storing the list in Bob’s safe) even before Alice has obtained her measurement results. And the outcome for B is defined before Alice has set her polarizer.
[...]

This is non-local. Bell's definition of locality in his 1964 paper is "that the result of a measurement on one system be unaffected
by operations on a distant system with which it has interacted in the past". This is clearly violated by your model, since you write "0.5 seconds after the peer photon 1 has left the earth Alice sets her polarizer to α0 and sends this information to Bob via a classical channel. After Bob gets this information he opens the safe and picks the outcome from the list for the submitted value of α0. A match occurs if this outcome is +1".

Since Bob's result depends on the value α0, this means that the result of the measurement depends on the operations on the distant system. It is irrelevant whether this happens superluminally or classically. There should be no communication whatsoever between Alice and Bob's stations for the locality condition to hold.

[Esail]

Bob measures a photon 2 having the parameter lambda0.

Since Bob's result depends on the value α0, this means that the result of the measurement depends on the operations on the distant system. It is irrelevant whether this happens superluminally or classically. There should be no communication whatsoever between Alice and Bob's stations for the locality condition to hold.

Implicitly I assumed we only regard Bob's result of +1 for the list in the safe as we want to detect matches (A=1,B=1). These are the events detected behind the polarizer set to beta. (I should add this to the story). So Bob's result does not depend on α0.
For (A=-1,B=-1) we can do the same procedure with the same result.

[Heinera]

Implicitly I assumed we only regard Bob's result of +1 for the list in the safe as we want to detect matches (A=1,B=1). These are the events detected behind the polarizer set to beta. (I should add this to the story). So Bob's result does not depend on α0.

Then why must Alice send the value of α0 to Bob?

[Esail]

Implicitly I assumed we only regard Bob's result of +1 for the list in the safe as we want to detect matches (A=1,B=1). These are the events detected behind the polarizer set to beta. (I should add this to the story). So Bob's result does not depend on α0.

Then why must Alice send the value of α0 to Bob?

Bobs list reads for example
α - possible result for A
0° +1
10° +1
20° +1
30° -1
40° -1
... ...

Setting α= α0 transmitted by Alice Bob gets the results for the chosen α0 (A=+1, B=+1) or (A=-1, B=+1)
The list can be replaced by a formula.
As we have seen Bob doesn't need Alices setting to obtain his results. After Bob's measurement the value of lambda for the pair is known. So Alice's result is determined given her setting α0 without any communication from Bob needed.

[Justo]

It is a quite simple matter to decide its locality. Lambada is supposed to be fixed at the source before any measurement is made. If Alice measures her photon, Bob's result not only depends on Lambda and his setting but also on Alice's setting. There is nothing more to it. No matter what words are used to justify that behavior, be it contextuality, indistinguishability, or whatever; the model is nonlocal.

[Heinera]

As we have seen Bob doesn't need Alices setting to obtain his results.

No, we have not seen that. If it is true, why does Alice need to send α0 to Bob?

[Esail]

As we have seen Bob doesn't need Alices setting to obtain his results.

No, we have not seen that. If it is true, why does Alice need to send α0 to Bob?

Bob's result =1 as we have seen. Alice does not need to send her result to Bob for the purpose of measurement. The purpose in the story was to bring informations from both side together without disturbing the measurement. The result is determined with the setting of alpha0 regardless if Bob knows it or not.

[Esail]

It is a quite simple matter to decide its locality. Lambada is supposed to be fixed at the source before any measurement is made. If Alice measures her photon, Bob's result not only depends on Lambda and his setting but also on Alice's setting. There is nothing more to it. No matter what words are used to justify that behavior, be it contextuality, indistinguishability, or whatever; the model is nonlocal.

If Alice detects her photon 1 at alpha then all photons selected by polarizer set to alpha have polarization alpha. Matching events occur for those photons 1 with polarization alpha which would hit a polarizer at side 1 set to beta-pi/2. Those photon 1 have a peer photon 2 which would hit polarizer PB set to beta. This is a local effect as no communication between side A and side be takes place for it. The reason for the same behavior are the same physical conditions at both sides: The same value of lambda and the same value of delta.
So we get A(delta, lambda) = B(delta, lambda)

[Justo]

If Alice detects her photon 1 at alpha then all photons selected by polarizer set to alpha have polarization alpha. Matching events occur for those photons 1 with polarization alpha which would hit a polarizer at side 1 set to beta-pi/2. Those photon 1 have a peer photon 2 which would hit polarizer PB set to beta. This is a local effect as no communication between side A and side be takes place for it. The reason for the same behavior are the same physical conditions at both sides: The same value of lambda and the same value of delta.
So we get A(delta, lambda) = B(delta, lambda)

Fine, now write the expression for delta. I understand it could depend on the case and there may be different cases. Do it for one particular case. For instance, let us say Alice measures his photon first, it is either 0 or 90 polarized. Let's it is 0 and alpha and beta given numbers. What is the expression for delta? If you want to convince us you should answer, besides its very easy for you.

[Esail]

If Alice detects her photon 1 at alpha then all photons selected by polarizer set to alpha have polarization alpha. Matching events occur for those photons 1 with polarization alpha which would hit a polarizer at side 1 set to beta-pi/2. Those photon 1 have a peer photon 2 which would hit polarizer PB set to beta. This is a local effect as no communication between side A and side be takes place for it. The reason for the same behavior are the same physical conditions at both sides: The same value of lambda and the same value of delta.
So we get A(delta, lambda) = B(delta, lambda)

Fine, now write the expression for delta. I understand it could depend on the case and there may be different cases. Do it for one particular case. For instance, let us say Alice measures his photon first, it is either 0 or 90 polarized. Let's it is 0 and alpha and beta given numbers. What is the expression for delta? If you want to convince us you should answer, besides its very easy for you.

With polarizer setting alpha at side A and beta at side B we have an arbitrary context (see the paper). This means (MA3) the polarization of Alice's photons 1 which are selected by her polarizer have polarization alpha. With MA4 delta is polarization - polarizer setting = alpha-alpha=0. This means A=1 for all photons 1 detected by this polarizer.

On the other side 2 the peer photon 2 of photon 1 have polarization alpha+pi/2 (see the paper): For a polarizer setting beta on side B we obtain
delta = alpha+pi/2 - beta. Here B depends on the value of lambda. Matching events occur if B=1. This is the case for 0<=delta<pi/2 if
lambda < cos(delta)*cos(delta).

[Justo]

For a polarizer setting beta on side B we obtain delta = alpha+pi/2 - beta. Here B depends on the value of lambda.

Yes, it depends on the value of lambda but it also depends on the value of alpha and beta. A local function for Bob can only depend on beta and lambda. If it depends on alpha(Alice setting) it cannot be a local function.
That is what everybody here and also in ResearchGate is trying to tell you. At last, thanks to you, Bell believers(like Richard and me) and Bell deniers(like Joy and Karl) agree on something. Congratulations!.

[Joy Christian]

At last, thanks to you, Bell believers(like Richard and me) and Bell deniers(like Joy and Karl) agree on something. Congratulations!.

Just for the record, Justo, serious researchers like Karl and I have never challenged Bell's mathematical formulation of what local means. His formulation is simply a mathematically precise version of Einstein's conception of locality in the present context. You will find that several other Bell-challengers in this forum also agree with Bell what local means. Our disagreement lies elsewhere, as both Karl, I, and others have explained in our respective publications many times before. But you are right that all parties agree that the model published in EPL is not local.
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[FrediFizzx]

... At last, thanks to you, Bell believers(like Richard and me) and Bell deniers(like Joy and Karl) agree on something. Congratulations!.

Justo, you and Gill are not just believers, you are Bell fanatics. Bell has been thoroughly shot down in more than one way. It is unfortunate (for you all) that you don't want to accept the truth of the matter and admit that Bell was wrong.
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[Gordon Watson]

At last, thanks to you, Bell believers(like Richard and me) and Bell deniers(like Joy and Karl) agree on something. Congratulations!.

Just for the record, Justo, serious researchers like Karl and I have never challenged Bell's mathematical formulation of what local means. His formulation is simply a mathematically precise version of Einstein's conception of locality in the present context. You will find that several other Bell-challengers in this forum also agree with Bell what local means. Our disagreement lies elsewhere, as both Karl, I, and others have explained in our respective publications many times before. But you are right that all parties agree that the model published in EPL is not local.
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Joy, what do you mean by this: "Bell's mathematical formulation of what local means."

In my view, Bell-local is not the same as Einstein-local.
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