Gordon Watson wrote:.
Bell's theorem has been described as the most profound discovery of science, one of the few essential discoveries of 20th Century physics, indecipherable to non-mathematicians. Let's see.
https://vixra.org/abs/2010.0068
ALL comments welcome, especially those that are educative and/or critical.
Thanks; Gordon
.
gill1109 wrote:Gordon Watson wrote:.
Bell's theorem has been described as the most profound discovery of science, one of the few essential discoveries of 20th Century physics, indecipherable to non-mathematicians. Let's see.
https://vixra.org/abs/2010.0068
ALL comments welcome, especially those that are educative and/or critical.
Thanks; Gordon
.
Just two pages! Excellent.
Q1. Notation. Why do you have “beta” in your notation, standing for “Bell’s experiment”?
Q2. Doesn’t every P(... | ...) in the whole two pages stand for probability in Bell’s experiment?
Q3. What do you mean by A^+, A^- and A^ +- ; similarly for B. Bell calls the actually observed results A and B; their possible values are +1 and -1.
Gordon Watson wrote:gill1109 wrote:Gordon Watson wrote:.
Bell's theorem has been described as the most profound discovery of science, one of the few essential discoveries of 20th Century physics, indecipherable to non-mathematicians. Let's see.
https://vixra.org/abs/2010.0068
ALL comments welcome, especially those that are educative and/or critical.
Thanks; Gordon
Just two pages! Excellent.
Q1. Notation. Why do you have “beta” in your notation, standing for “Bell’s experiment”?
Q2. Doesn’t every P(... | ...) in the whole two pages stand for probability in Bell’s experiment?
Q3. What do you mean by A^+, A^- and A^ +- ; similarly for B. Bell calls the actually observed results A and B; their possible values are +1 and -1.
A1. β (beta) denotes the experiment IN Bell (1964). That experiment is the EPR-Bohm experiment. Elsewhere I chose β to honour David Bohm. In trying to keep to 2 pages here, I omitted such background.
A2. NO; see the line before B(2): Bell uses P to denote an expectation value. Further: nowhere does Bell use this crucial bit in any probability function -- |...) -- whereas you rightly show it in your notation above.
A3. Please note: Bell is confused here; see the line before B(1)! He uses A and B for both functions and the related results. Since the related results are +1 or -1, it is much clearer to use my terms in probability relations. See my move from eqn (3) -- with its functions -- to eqn (4) with its more conventional [encoded, and therefore more compact] results.
HTH; Gordon
gill1109 wrote:Gordon Watson wrote:gill1109 wrote:Gordon Watson wrote:.
Bell's theorem has been described as the most profound discovery of science, one of the few essential discoveries of 20th Century physics, indecipherable to non-mathematicians. Let's see.
https://vixra.org/abs/2010.0068
ALL comments welcome, especially those that are educative and/or critical.
Thanks; Gordon
Just two pages! Excellent.
Q1. Notation. Why do you have “beta” in your notation, standing for “Bell’s experiment”?
Q2. Doesn’t every P(... | ...) in the whole two pages stand for probability in Bell’s experiment?
Q3. What do you mean by A^+, A^- and A^ +- ; similarly for B. Bell calls the actually observed results A and B; their possible values are +1 and -1.
A1. β (beta) denotes the experiment IN Bell (1964). That experiment is the EPR-Bohm experiment. Elsewhere I chose β to honour David Bohm. In trying to keep to 2 pages here, I omitted such background.
A2. NO; see the line before B(2): Bell uses P to denote an expectation value. Further: nowhere does Bell use this crucial bit in any probability function -- |...) -- whereas you rightly show it in your notation above.
A3. Please note: Bell is confused here; see the line before B(1)! He uses A and B for both functions and the related results. Since the related results are +1 or -1, it is much clearer to use my terms in probability relations. See my move from eqn (3) -- with its functions -- to eqn (4) with its more conventional [encoded, and therefore more compact] results.
HTH; Gordon
Just trying to be educative, but also trying to make the job of reading the two pages more pleasant:
A1. The whole paper assumes the EPR-Bohm experiment as background. You can safely omit beta *everywhere*. Removing superfluous letters on the pages would make them *much* easier to read.
A2. Yes, I know you are using better notation than Bell, in having E for expectation. That’s not my point at all.
A3. You can use different letters, or different letter-styles (Italic/Roman/Gothic...; bold, double, ...) or different alphabets (Greek/Hebrew/...) or special diacritical marks (hats, bars, ...) for functions , and for outcomes - possible values taken by functions. Your superscript +, -, and +- are superfluous and confusing.
Just delete every symbol which is superfluous, and the whole thing will be *much* easier to read. Less is more!
Then on to a new question:
Q4. Inside an integration, and apparently to be interpreted as a number multiplying another number, you write “(A(a, lambda) = 1)”. But the result of evaluating “A(a, lambda) = 1” is true or false. Are you wanting us to convert that to 1 or 0, accordingly?
Gordon Watson wrote:Richard, etc.,
Seeking to be clearer: here's Version 2, https://vixra.org/pdf/2010.0068v2.pdf
In eqn (3), to identify each integrand, I've this time used (the triple bar) to denote equivalence: and thereby separate the results.
I personally prefer for equivalence.
So maybe this would be better; and even clearer?
and ?
Thanks; Gordon
gill1109 wrote:Gordon Watson wrote:Richard, etc.,
Seeking to be clearer: here's Version 2, https://vixra.org/pdf/2010.0068v2.pdf
In eqn (3), to identify each integrand, I've this time used (the triple bar) to denote equivalence: and thereby separate the results.
I personally prefer for equivalence.
So maybe this would be better; and even clearer?
and ?
Thanks; Gordon
It looks to me this is overkill. You say “something” is in the set of x such that x is equivalent to 1. So you could replace that by just saying that “something” is equivalent to 1. And what does “equivalent” mean? Don’t you mean “equals”?
Gordon Watson wrote:gill1109 wrote:Gordon Watson wrote:Richard, etc.,
Seeking to be clearer: here's Version 2, https://vixra.org/pdf/2010.0068v2.pdf
In eqn (3), to identify each integrand, I've this time used (the triple bar) to denote equivalence: and thereby separate the results.
I personally prefer for equivalence.
So maybe this would be better; and even clearer?
and ?
Thanks; Gordon
It looks to me this is overkill. You say “something” is in the set of x such that x is equivalent to 1. So you could replace that by just saying that “something” is equivalent to 1. And what does “equivalent” mean? Don’t you mean “equals”?
Thanks Richard,
Given your earlier comments, I thought my earlier use of "equals" could be improved. (For you seemed to suggest it was some unique form of mathematics.)
Equivalence is a more general term than equals. So we are separating those AB functions (with a λ-pair that delivers 1) from those AB functions (with a λ-pair that delivers -1).
That way you should not be confused into thinking that the integrands can be reduced to ±1 before integrating. And in this way -- under the "overkill" -- the integral in (1) reduces to eqn (4) rigorously.
Richard, in that you seemed to recently think that Bell's use of P denoted probabilities, why not begin with the refutation of Bell's inequality that I now include in the 2-page format that you favoured? For if that simple analysis is false ... I'm out of here!
That way, in the interim, those who understand set-notation can assist with the following.
I think this is the best more-expansive way to go, thus far, in my eqn (3):
and ?
.
gill1109 wrote:Gordon Watson wrote:gill1109 wrote:Gordon Watson wrote:Richard, etc.,
Seeking to be clearer: here's Version 2, https://vixra.org/pdf/2010.0068v2.pdf
In eqn (3), to identify each integrand, I've this time used (the triple bar) to denote equivalence: and thereby separate the results.
I personally prefer for equivalence.
So maybe this would be better; and even clearer?
and ?
Thanks; Gordon
It looks to me this is overkill. You say “something” is in the set of x such that x is equivalent to 1. So you could replace that by just saying that “something” is equivalent to 1. And what does “equivalent” mean? Don’t you mean “equals”?
Thanks Richard,
Given your earlier comments, I thought my earlier use of "equals" could be improved. (For you seemed to suggest it was some unique form of mathematics.)
Equivalence is a more general term than equals. So we are separating those AB functions (with a λ-pair that delivers 1) from those AB functions (with a λ-pair that delivers -1).
That way you should not be confused into thinking that the integrands can be reduced to ±1 before integrating. And in this way -- under the "overkill" -- the integral in (1) reduces to eqn (4) rigorously.
Richard, in that you seemed to recently think that Bell's use of P denoted probabilities, why not begin with the refutation of Bell's inequality that I now include in the 2-page format that you favoured? For if that simple analysis is false ... I'm out of here!
That way, in the interim, those who understand set-notation can assist with the following.
I think this is the best more-expansive way to go, thus far, in my eqn (3):
and ?
.
I did not think Bell’s use of P denoted probability.
A times B is one function. You can’t “separate those functions which deliver this, or that”. It makes no sense at all. You can separate their arguments. Or you can create new functions by writing e’g. f = f 1_A + f 1_Ac. Write f as a sum of two parts, each zero on complementary parts of their domain; I use 1_A to denote the indicator function of the set A. It takes the values 1 and 0 depending on whether x in A or not.
gill1109 wrote:I would use
Z: AB (AB == 1) ,
having explained that in an arithmetical expression, logical expressions are converted from T/F to 1/0. I would use == when asking are LHS and RHS equal, forcing AB == 1 to be read as logic. I would use := for defining LHS by RHS, and plain = for a chain of expressions asserting equality.
In Z, three functions are multiplied together: A, B, and (AB == 1). The third function is a function of a, b and lambda and it takes values in {0, 1}.
A(a, lambda) is a number
A is a function
You did not define “x is equivalent to 1”, and the set of numbers x equal to 1 contains only one number.
So X and Y are awful. Say simple things in a simple way. Mathematical notation is designed in order to say difficult things as clearly and simply as possible. Mathematics is built up logically and sequentially. Every time you introduce new notation, or you change notation, you should say so. Usually you are writing in a context which means that you know the notation which your readers are familiar with, and you use it too, if possible, or at least, to start with. You build on foundations which are there and which are known by your readers.
I realise how difficult it must be! I have had these discussions with students for 45 years and once I was a student myself. I remember a student asking me (after the lecture) “does the less than or equals symbol” mean less than or equals, or less than and equals? I wish he had dared to ask it during the lecture.
gill1109 wrote:Bell argues that functions A, B and rho do not exist such that E(a, b) := int d lambda rho(lambda) A(a, lambda) B(b, lambda) = - a.b for all a and b, where the functions A and B take values in the set {-1, +1}; a and b are arbitrary directions in space; lambda lies in some space of hidden variables Lambda, and rho is a probability density over that space (non-negative, integrates to 1). Naturally, one can write the integral as the sum of two integrals, one over the region where A(a, .)B(b, .) = +1, and one over the complementary region where A(a, .)B(b, .) = -1, and hence find that E(a, b) = P(A_a B_b = +1) - P(A_a B_b = -1), where we now consider Lambda together with the probability density rho as defining a probability space with probability measure P; we can now define A_a and B_b as the random variables (functions of lambda) A(a, .) and B(b, .).
Your notation is confusing since the bar "|" is usually used to mean "conditional on" in probability theory. If you want to integrate over subsets of lambda then you can indicate that by giving the integral sign a subscript (over which set do you integrate); or you can multiply the integrand by an indicator function. In your last line, you have on the right-hand side dropped the variables a and b. You write a lot of stuff which is superfluous. The short form "AB" for a random variable which also depends on a and b is dangerous and misleading. The A^+- is quite superfluous. beta is superfluous. Bell is assuming local realism. The "EPR-Bohm set-up" is irrelevant. You are going to do some mathematics, not physics.
Indeed the bar is also used in "set-builder" notation. But you are not using it to build sets. A(a, lambda)B(b, lambda) is a number. {lambda | A(a, lambda)B(b, lambda) = +1} is a set, and moreover, it is a set which in general might depend on a and b. But I don't see any curly brackets, so I don't see any set-builder sets, and anyway, one integrates *functions* over sets. You are mixing up functions and sets and this kind of mix-up is a recipe for disaster, especially when at the same time you put many symbols into your notation which are superfluous but at the same time omit symbols which are important. You end up writing something which makes no sense - there are a and b on the left-hand side, but not on the right-hand side.
All of which still leaves you needing to identify an error in my refutation of BT.
Gordon Watson wrote:With my emphasis (to identify bits that confuse me):gill1109 wrote:Bell argues that functions A, B and rho do not exist such that E(a, b) := int d lambda rho(lambda) A(a, lambda) B(b, lambda) = - a.b for all a and b, where the functions A and B take values in the set {-1, +1}; a and b are arbitrary directions in space; lambda lies in some space of hidden variables Lambda, and rho is a probability density over that space (non-negative, integrates to 1). Naturally, one can write the integral as the sum of two integrals, one over the region where A(a, .)B(b, .) = +1, and one over the complementary region where A(a, .)B(b, .) = -1, and hence find that E(a, b) = P(A_a B_b = +1) - P(A_a B_b = -1), where we now consider Lambda together with the probability density rho as defining a probability space with probability measure P; we can now define A_a and B_b as the random variables (functions of lambda) A(a, .) and B(b, .).
Your notation is confusing since the bar "|" is usually used to mean "conditional on" in probability theory. If you want to integrate over subsets of lambda then you can indicate that by giving the integral sign a subscript (over which set do you integrate); or you can multiply the integrand by an indicator function. In your last line, you have on the right-hand side dropped the variables a and b. You write a lot of stuff which is superfluous. The short form "AB" for a random variable which also depends on a and b is dangerous and misleading. The A^+- is quite superfluous. beta is superfluous. Bell is assuming local realism. The "EPR-Bohm set-up" is irrelevant. You are going to do some mathematics, not physics.
Indeed the bar is also used in "set-builder" notation. But you are not using it to build sets. A(a, lambda)B(b, lambda) is a number. {lambda | A(a, lambda)B(b, lambda) = +1} is a set, and moreover, it is a set which in general might depend on a and b. But I don't see any curly brackets, so I don't see any set-builder sets, and anyway, one integrates *functions* over sets. You are mixing up functions and sets and this kind of mix-up is a recipe for disaster, especially when at the same time you put many symbols into your notation which are superfluous but at the same time omit symbols which are important. You end up writing something which makes no sense - there are a and b on the left-hand side, but not on the right-hand side.
Thanks Richard, very helpful.
Please see if the version just posted meets your needs and allows you to pinpoint my errors.
Note:
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