Commonsense local realism refutes Bell's theorem

[gill1109]

Some references:

http://arxiv.org/abs/1402.1972
Constraints on determinism: Bell versus Conway-Kochen
Eric Cator, Klaas Landsman
http://www.math.leidenuniv.nl/~gill/Klaas_Landsman_Bell_talk.pdf

http://en.citizendium.org/wiki/entanglement_(physics)
Tsirelson's article on Citizendium

[minkwe]

The expression above can also be written as

E(a, b) = integral A(a, lambda) B(b, lambda) rho(a, b, lambda) d lambda

Now it is easier to see that it is LHV-specific, albeit with a conspiracy term rho(a, b, lambda).

Heine, this escape will not work. A conspiracy theory is one in which Alice and Bob conspire to use each other's settings. If Alice and Bob do not know each others settings, the theory can not be conspiracy, otherwise please tell us what the conspiracy is. You had a foot-in-mouth problem claiming that the expression was only well defined for LHV theories now you want to have a bullet-in-foot problem by claiming that the expression

E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b)

must be a conspiracy theory. That is simply not true as can be see by simply "thinking" about it. Alice and Bob do not know what angle each other is going to use. They are randomly selecting their angle every time they make a measurement. Yet each time they make a choice of angle, the choice is instantaneously transmitted to the other side and used together with the other person's choice to determine the outcome there. There is no conspiracy here. It is blatant non-locality, and the integral is well defined, and has exactly the same functional form. You were wrong, stop digging already. There is nothing LHV-specific about the integral. It is only LHV specific if you define lambda as a shared property which could only have been shared through an interaction in the common region of their light-cones. You can simply change the meaning of lambda into a property shared instantaneously and the model becomes non-local with exactly the same integral still applying. I just gave you one simple example which lambda = (a,b) and that sunk your claims.

[minkwe]

Michel, did you read the later papers of John Bell yet?

Richard, have you found anything yet in any of Bell's latest papers which rescues his fatal mixing of weakly and strongly objective expectation values? When you find it, please present it at viewtopic.php?f=6&t=63

[gill1109]

Michel, did you read the later papers of John Bell yet?

Richard, have you found anything yet in any of Bell's latest papers which rescues his fatal mixing of weakly and strongly objective expectation values? When you find it, please present it at viewtopic.php?f=6&t=63

I wonder, Michel, if you so appreciate Jaynes, why you don't use Jaynes' Bayesian notion of probability.

I have no use whatsoever for a "strongly objective" interpretation of probability. Moreover we have to carefully separate the worlds of reality, of mathematics, and the bridges between them. The bridges are interpretations. In particular, interpretations of probability. Bridges between physical reality and mathematical models are what we call metaphysics.

Bell is not mixing anything up, but both you and Gordon Watson are badly mixing up both levels and categories. Categories: name versus value, for instance. Categories: physical reality versus mathematical model. Levels: you are mixing up the bridges between worlds and the objects inside those worlds.

In order to do meta-physics, one needs to take a higher view somehow "above" both mathematics and physics, separately.

[Heinera]

The expression above can also be written as

E(a, b) = integral A(a, lambda) B(b, lambda) rho(a, b, lambda) d lambda

Now it is easier to see that it is LHV-specific, albeit with a conspiracy term rho(a, b, lambda).

Heine, this escape will not work. A conspiracy theory is one in which Alice and Bob conspire to use each other's settings. If Alice and Bob do not know each others settings, the theory can not be conspiracy, otherwise please tell us what the conspiracy is.

The "conspiracy" is meant to be a conspiracy of nature; since the generation of hidden variables depending on settings a and b takes place before these are decided by Alice and Bob, nature must have predetermined these in advance. Also known as superdeterminism.

You had a foot-in-mouth problem claiming that the expression was only well defined for LHV theories now you want to have a bullet-in-foot problem by claiming that the expression

E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b)

must be a conspiracy theory.

I did not say that. I say that the expression is meaningless. On left hand side you have two free variables a and b, on the right hand side you have none, since you integrate over (a,b).

[minkwe]

The "conspiracy" is meant to be a conspiracy of nature; since the generation of hidden variables depending on settings a and b takes place before these are decided by Alice and Bob, nature must have predetermined these in advance. Aka known as superdeterminism.

I can also say non-locality is a conspiracy of nature, and LHV is a conspiracy of nature. Like I said, stop digging already.

I did not say that. I say that the expression is meaningless.

Wrong. For a non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined.

Just admit that you were wrong and we'll leave it at that. the functions A(a, lambda) and B(b, lambda) are well defined for non-local theories contrary to your claims. At least stop digging.

[Heinera]

The "conspiracy" is meant to be a conspiracy of nature; since the generation of hidden variables depending on settings a and b takes place before these are decided by Alice and Bob, nature must have predetermined these in advance. Aka known as superdeterminism.

I can also say non-locality is a conspiracy of nature, and LHV is a conspiracy of nature. Like I said, stop digging already.

Of course you can. You can define the word "conspiracy" any way you like. But if you want people to understand you, it will be easier to use a definition roughly equal to what other people use.

I did not say that. I say that the expression is meaningless.

Wrong. For a non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined.

Just admit that you were wrong and we'll leave it at that. the functions A(a, lambda) and B(b, lambda) are well defined for non-local theories contrary to your claims. At least stop digging.

Why should I admit that? Any attempt at incorporating (a,b) into the hidden variable will lead to the meaningless equation E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b).

[minkwe]

Why should I admit that? Any attempt at incorporating (a,b) into the hidden variable will lead to the meaningless equation E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b).

You claimed that "For non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined". That has been shown to be a lie, you can't even bring yourself to admit it, now you are rambling about integrals.

Let lambda be a non-local hidden variable, please write down the integral for the expectation value of the paired-product of outcomes at Alice and Bob. [Hint: it is exactly the same as Bell's equation 2, the one you claim only applies to local-hidden variable theories]

[Heinera]

Why should I admit that? Any attempt at incorporating (a,b) into the hidden variable will lead to the meaningless equation E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b).

You claimed that "For non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined". That has been shown to be a lie, you can't even bring yourself to admit it, now you are rambling about integrals.

Where has that been shown to be a "lie"?

[minkwe]

Why should I admit that? Any attempt at incorporating (a,b) into the hidden variable will lead to the meaningless equation E(a, b) = integral A(a, b) B(b, b) rho(a, b) d (a,b).

You claimed that "For non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined". That has been shown to be a lie, you can't even bring yourself to admit it, now you are rambling about integrals.

Where has that been shown to be a "lie"?

if lambda = (a,b), then A(a, lambda) = A(a,b) and B(b, lambda) = B(a,b)

Both functions A(a,b) and B(a,b) are well defined. Therefore you were wrong.

If lambda is a non-local hidden variable, then the functions A(a, lambda), and B(b, lambda) are well defined. There is nothing LHV-specific in those functions. They are simply functions. lambda can be a local hidden variable, or a non-local hidden variable or a conspiracy variable or any variable you like. The physical properties you attribute to lambda, does not change the fact that the functions A(a, lambda), and B(b, lambda) are well defined contrary to your claims. Even your own non-local model which you presented in response to my earlier challenge as a non-local model, has well defined A(a, lambda), and B(b, lambda) functions, so you knew that it was well defined but claimed otherwise.

Let lambda be a non-local hidden variable, please write down the integral for the expectation value of the paired-product of outcomes at Alice and Bob. [Hint: it is exactly the same as Bell's equation 2, the one you claim only applies to local-hidden variable theories]

[Heinera]

You claimed that "For non-local theory, the functions A(a, lambda) and B(b, lambda) are not even well defined". That has been shown to be a lie, you can't even bring yourself to admit it, now you are rambling about integrals.

Where has that been shown to be a "lie"?

if lambda = (a,b), then A(a, lambda) = A(a,b) and B(b, lambda) = B(a,b)

Both functions A(a,b) and B(a,b) are well defined. Therefore you were wrong.

Even your own non-local model which you presented in response to my earlier challenge as a non-local model, has well defined A(a, lambda), and B(b, lambda) functions, so you knew that it was well defined but claimed otherwise.

No. In that model neither A(a, lambda) nor B(b, lambda) has well defined values; they both also depend (via global variables) on the other party's b and a, respectively (so e.g A(a, lambda) can have multiple values for the same arguments (a, lambda), depending upon b). This means that A(a, lambda), and B(b, lambda) are not mathmatically well defined as functions for those arguments. They can only mathematically correctly be written A(a, b, lambda) and B(a, b, lambda), and then they are of course completely outside the scope of Bell's theorem (which is why that simulation can do what it does)

[minkwe]

No. In that model neither A(a, lambda) nor B(b, lambda) has well defined values; they both also depend on b and a, respectively (so they can have multiple values for the same arguments (a, lambda) and (b,lambda)

In other words, your lambda = (a,b,hv), and your functions A(a,lambda), B(b, lambda) are well defined. A(a,hv) is not well defined but A(a,lambda) is. You can't escape the point that for non-local hidden variables A(a,lambda) is well defined by simply defining your lambda to exclude the non-local variables required by the function. It won't work.

Let lambda be a non-local hidden variable, please write down the integral for the expectation value of the paired-product of outcomes at Alice and Bob.

[Heinera]

No. In that model neither A(a, lambda) nor B(b, lambda) has well defined values; they both also depend on b and a, respectively (so they can have multiple values for the same arguments (a, lambda) and (b,lambda)

In other words, your lambda = (a,b,hv), and your functions A(a,lambda), B(b, lambda) are well defined.

In the model we are talking about, the variable lambda is a random number between -1 and 1. If you want to create a new model where the varaiable lambda includes the detector settings (a,b), fine. In that model Bell's proof will break down (we end up with a certain meaningless expression for E(a, b), as alreay discussed), so his theorem doesn't apply to this case either. So a HV model where the hidden variable actually consists of both detector settings is perfectly consistent, and in no way in conflict with Bell's theorem. Its called superdeterminism. Problem is, it's hard to sell.

[minkwe]

Let lambda be a non-local hidden variable, please write down the integral for the expectation value of the paired-product of outcomes at Alice and Bob.

When lambda is non-local hidden variable, What is wrong with




Please show exactly why the above three expressions are as you call it "undefined" or "meaningless" for a non-local hidden variable lambda. I've been asking you for a while, maybe one of your Bell-believer friends can help you explain what exactly in the above expression restricts it to LHV.

[Xray]

No. In that model neither A(a, lambda) nor B(b, lambda) has well defined values; they both also depend on b and a, respectively (so they can have multiple values for the same arguments (a, lambda) and (b,lambda)

In other words, your lambda = (a,b,hv), and your functions A(a,lambda), B(b, lambda) are well defined.

In the model we are talking about, the variable lambda is a random number between -1 and 1.

Hi Heinera

Does the variable lambda have units or does it stand for something else in your model because how does anything other than a come out of A(a, -1 <= lambda <= 1) and we need A(a, …) = ±1 to agree with Bell?

Xray

[FrediFizzx]

If a is a vector then lambda must have a vector in it to get a scalar for the final result.

[gill1109]

If a is a vector then lambda must have a vector in it to get a scalar for the final result.

A(a, lambda) is a function of setting and hidden variable. The hidden variable can lie in any space of any kind, large or small, you like.

[FrediFizzx]

Is the "setting" represented by a vector? If not, what is its definition?

[gill1109]

Is the "setting" represented by a vector? If not, what is its definition?

"Settings" are chosen by experimenters by pressing buttons or turning dials on pieces of apparatus. In mathematical physical theories they could be represented in all kinds of ways. Talking about EPR-B, they are pairs of directions in 3-d space.

[FrediFizzx]

Is the "setting" represented by a vector? If not, what is its definition?

"Settings" are chosen by experimenters by pressing buttons or turning dials on pieces of apparatus. In mathematical physical theories they could be represented in all kinds of ways. Talking about EPR-B, they are pairs of directions in 3-d space.

So vectors.