A new, simplified derivation of the EPR-Bohm correlation

[Joy Christian]

Perhaps the notation you are using on the RHS might be messing up some people? I think usually it is something like s --> (a', b') or would it be (s, s) --> (a', b')?
http://tutorial.math.lamar.edu/Classes/ ... imits.aspx

Anyways, it is perfectly understandable what you are doing either way.

It is a standard notation. But the above link is a good find. It is convenient to have here, in case someone is confused about my notation. No mathematician would be.

[FrediFizzx]

Perhaps the notation you are using on the RHS might be messing up some people? I think usually it is something like s --> (a', b') or would it be (s, s) --> (a', b')?
http://tutorial.math.lamar.edu/Classes/ ... imits.aspx

Anyways, it is perfectly understandable what you are doing either way.

It is a standard notation. But the above link is a good find. It is convenient to have here, in case someone is confused about my notation. No mathematician would be.

Yes, and the bottom line here with the new paper and even the old one is that the detector bivectors drop out of the correlation calculation as they should. However, with the realization that Bell's argument is "rigged" against both LHV and QM models as far as +/-1 outcomes go, you can just do,

(mu_k.a)(mu_k.b) = -a.b + 0,

and be done with it, IMHO. What is the point in trying to follow Bell's junk physics?

[Joy Christian]

Yes, and the bottom line here with the new paper and even the old one is that the detector bivectors drop out of the correlation calculation as they should. However, with the realization that Bell's argument is "rigged" against both LHV and QM models as far as +/-1 outcomes go, you can just do,

(mu_k.a)(mu_k.b) = -a.b + 0,

and be done with it, IMHO. What is the point in trying to follow Bell's junk physics?

Well, let us try to look for a silver lining in the dark cloud. Since even after 50 years of misdirection the physics community is still chasing a "spook" out of a Bell jar, that is the best we can do. The silver lining is that we now understand that EPR-Bohm correlations are correlations between the scalar points, A = +/-1 and B = +/-1, of a quaternionic 3-sphere, which is of course the spatial part of one of the most well known cosmological solutions of Einstein's field equations of general relativity:

[FrediFizzx]

Yes, and the bottom line here with the new paper and even the old one is that the detector bivectors drop out of the correlation calculation as they should. However, with the realization that Bell's argument is "rigged" against both LHV and QM models as far as +/-1 outcomes go, you can just do,

(mu_k.a)(mu_k.b) = -a.b + 0,

and be done with it, IMHO. What is the point in trying to follow Bell's junk physics?

Well, let us try to look for a silver lining in the dark cloud. Since even after 50 years of misdirection the physics community is still chasing a "spook" out of a Bell jar, that is the best we can do. The silver lining is that we now understand that EPR-Bohm correlations are correlations between the scalar points, A = +/-1 and B = +/-1, of a quaternionic 3-sphere, which is of course the spatial part of one of the most well known cosmological solutions of Einstein's field equations of general relativity:

Ah... yes. And the solution to that is here. So not really following Bell's R^3 junk physics.

[Joy Christian]

***
In addition to revising this one-page paper, I have now also added a new one-page appendix to my "Reply to Gill." For convenience, I reproduce the appendix here:





Note that, as before, logically the only new ingredient needed to derive the "quantum" correlation is that of a relationship between the spin bivectors and detector bivectors specified by Eq. (C3). This concept is elaborated in great detail on page 11 (see Definition V.1) of this earlier paper: http://arxiv.org/pdf/1211.0784.pdf.

***

[Gordon Watson]

Joy, I'm not across this stuff, and I'm here late, but I wonder: Should the middle term in eq. (C14) have a minus sign to the left. I ask because you use the same λ^k for the A and B results

Conservation of angular momentum suggests to me that Alice's λ^k would be the opposite of Bob's λ^k, at least in 3-space.

Thanks; G

[FrediFizzx]

Joy, I'm not across this stuff, and I'm here late, but I wonder: Should the middle term in eq. (C14) have a minus sign to the left. I ask because you use the same λ^k for the A and B results

Conservation of angular momentum suggests to me that Alice's λ^k would be the opposite of Bob's λ^k, at least in 3-space.

Thanks; G

There is no separate λ^k for A and B per particle pair in Joy's model. Angular momentum is conserved in eq. (C5).

[Joy Christian]

Joy, I'm not across this stuff, and I'm here late, but I wonder: Should the middle term in eq. (C14) have a minus sign to the left. I ask because you use the same λ^k for the A and B results

Conservation of angular momentum suggests to me that Alice's λ^k would be the opposite of Bob's λ^k, at least in 3-space.

Thanks; G

Fred has already answered your question, but let me spell it out more explicitly: The conservation of zero spin angular momentum in the model simply means



for the same (i.e., for the same run of the experiment). must be the same for a given run because that is the initial state of the system for that run.

[Joy Christian]

***
For convenience I have posted the above page from by "Reply to Gill" on Academia.edu: https://www.academia.edu/17783877/Dispr ... 501.03393_

***

[Joy Christian]

***
I have now also incorporated the above simplified derivation of the strong correlation in my "local causality" paper: viewtopic.php?f=6&t=55&p=5791#p5791

***