Joy Christian's LHV Model that disproves Bell - Printable Version +- Hidden Variables ( https://ilja-schmelzer.de/hidden-variables)+-- Forum: Foundations of Quantum Theory ( https://ilja-schmelzer.de/hidden-variables/forumdisplay.php?fid=3)+--- Forum: The Violation of Bell's Inequalities ( https://ilja-schmelzer.de/hidden-variables/forumdisplay.php?fid=7)+--- Thread: Joy Christian's LHV Model that disproves Bell ( /showthread.php?tid=55) |

RE: Joy Christian's LHV Model that disproves Bell - FrediFizzx - 06-16-2016
(06-16-2016, 01:48 AM)secur Wrote: It seems that he's representing spin states as quaternions; not sure of the details. Well, you need to go all the way through to eq. (44) for the full impact that Joy's model is a classical local-realistic model that exactly produces the predictions of QM for the EPR-Bohm scenario. IOW, it answers your questions about matching QM. The FLRW part is to support the \(S^3\) postulate of the model. Then I believe the quaternionic stuff is extracted from \(S^3\). Of course it is handy that it matches up to spin. ... RE: Joy Christian's LHV Model that disproves Bell - secur - 06-16-2016
Not motivated to read the whole thing; as far as I know it could be alright. Consider, QM itself uses no hidden variables, nor any explicit non-locality, it just postulates that the correct correlations happen. When we look at that logically we decide for those correlations to exist there must be non-local FTL connection between Alice and Bob measurements, in some unknown way. Of course can't be used for communication. At best, Christian's paper provides a different mechanism for those correlations. That's cool. The S3 postulate, and/or FLRW, in addition to giving quaternionic correlations, provides a curved spacetime between Alice and Bob. So I imagine that's how the connection happens. It might be local in that curved space - playing a role reminiscent of a wormhole -, but in the one we live in, we still must perceive it as non-local. Less interesting, there may be some hidden variable hiding in all those equations. But it doesn't invalidate Bell. He says that in order to explain the experiment you must introduce some new factor which is essentially equivalent to non-locality. People have already come up with about 5, such as MWI, determinism, denying counterfactual definiteness, "conspiracy" theories - and now this: a new approach to spin topology, "local FLRW" or S3. If correct it's worth adding to the list! At best it's an interesting wrinkle on this issue. At worst there may be mistakes in the paper, as others have alleged. Let others figure it out, and let me know. Thanks FrediFizzx ! RE: Joy Christian's LHV Model that disproves Bell - FrediFizzx - 06-16-2016
(06-16-2016, 04:40 AM)secur Wrote: Not motivated to read the whole thing; as far as I know it could be alright. Yes, you missed the hidden variable. It is that the particle pairs as a system can be either left or right hand oriented. Too bad you don't seem to care about a full understanding because it is quite profound and leads to new physics. Everything in the model is local and realistic so it does invalidate Bell's theory. However, it does not follow Bell's "rules" because he made a mistake in such. He didn't realize that you can't put \(S^3\) in as a hidden variable. ... RE: Joy Christian's LHV Model that disproves Bell - Schmelzer - 06-16-2016
Of course, it does not invalidate Bell's theorem, because it is simply a construction which has nothing to do with the theorem. It constructs some functions \(A(a,\lambda), B(b,\lambda)\in S^3\) while in Bell's theorem one needs \(A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2\). RE: Joy Christian's LHV Model that disproves Bell - FrediFizzx - 06-16-2016
(06-16-2016, 06:38 AM)Schmelzer Wrote: Of course, it does not invalidate Bell's theorem, because it is simply a construction which has nothing to do with the theorem. It constructs some functions \(A(a,\lambda), B(b,\lambda)\in S^3\) while in Bell's theorem one needs \(A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2\).... It is quite curious that you want to stick to that nonsense. Bell's theorem is just a word statement. I will quote from the Wikipedia article on Bell's theorem. "Bell himself wrote: "If [a hidden variable theory] is local it will not agree with quantum mechanics, and if it agrees with quantum mechanics it will not be local. This is what the theorem says." John Bell, Speakable and Unspeakable in Quantum Mechanics, Cambridge University Press, 1987, p. 65" We have a classical local-realistic theory that does agree with QM so that statement is definitely wrong. ... RE: Joy Christian's LHV Model that disproves Bell - secur - 06-16-2016
FrediFizzx: Yes, you missed the hidden variable. It is that the particle pairs as a system can be either left or right hand oriented. Alright. FrediFizzx: Too bad you don't seem to care about a full understanding because it is quite profound and leads to new physics. The main problem is, there seems no motivation at all for this different topology. But more important, I'm working on profound new physics already - much more significant than this. Furthermore it will do me no good at all, except intellectual satisfaction. No credit, no money, no nothing. The very best I can hope for is: someone who is "in" steals my ideas, painting me as some whacko who tried to steal the ideas from him. Profound new physics is not all it's cracked up to be. As a general rule, taking a nice nap or playing with a grandchild is far more important than all the profound new physics the world has to offer. --------- FrediFizzx: Everything in the model is local and realistic so it does invalidate Bell's theory. However, it does not follow Bell's "rules" because he made a mistake in such. He didn't realize that you can't put \(S^3\) in as a hidden variable. Schmelzer: Of course, it does not invalidate Bell's theorem, because it is simply a construction which has nothing to do with the theorem. It constructs some functions A(a,λ),B(b,λ)∈S 3 A(a,λ),B(b,λ)∈S3 A(a,\lambda), B(b,\lambda)\in S^3 while in Bell's theorem one needs A(a,λ),B(b,λ)∈Z 2 A(a,λ),B(b,λ)∈Z2 A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2 . Schmelzer is of course right, but maybe Christian can come up with some justification for substituting this new topology. If so it becomes a sixth alternative for avoiding non-locality. Bell didn't "miss" anything. He allowed for any technique that would do the equivalent job as non-locality: like denying reality, denying counterfactual definiteness, and - maybe - claiming that QM spin space is really S3 instead of Z2. For all I know that might work, but if so, it's just another possibility like MWI. To me, simple non-locality is easier to understand and unobjectionable. Still if Christian wants to make that claim and can somehow make sense of it, good luck to him. It's worth doing - unless you have a grandchild to play with RE: Joy Christian's LHV Model that disproves Bell - gill1109 - 06-18-2016
(06-16-2016, 06:38 AM)Schmelzer Wrote: Of course, it does not invalidate Bell's theorem, because it is simply a construction which has nothing to do with the theorem. It constructs some functions \(A(a,\lambda), B(b,\lambda)\in S^3\) while in Bell's theorem one needs \(A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2\). In equations (1) and (2) of http://arxiv.org/pdf/1103.1879v2.pdf Christian points out that the functions he has constructed *do* satisfy \(A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2\). Trouble is, he makes a mistake computing the correlation. RE: Joy Christian's LHV Model that disproves Bell - Schmelzer - 06-18-2016
Ok, he makes different errors in different papers. RE: Joy Christian's LHV Model that disproves Bell - FrediFizzx - 06-18-2016
(06-18-2016, 02:59 PM)gill1109 Wrote:...(06-16-2016, 06:38 AM)Schmelzer Wrote: Of course, it does not invalidate Bell's theorem, because it is simply a construction which has nothing to do with the theorem. It constructs some functions \(A(a,\lambda), B(b,\lambda)\in S^3\) while in Bell's theorem one needs \(A(a,\lambda), B(b,\lambda)\in \mathbb{Z}_2\). Again, that is pure nonsense. The functions A and B are in \(S^3\). Bell didn't allow for \(S^3\) in his "rules"; you can't put \(S^3\) in as a hidden variable . So you have to do the correlation calculation how Joy did it in eqs. (5-12). If you guys want to stay stuck in "flatland", I have no problem with that. But at least tell the truth. All you guys are really doing is rejecting the postulate(s). That is fine if you wish, but there are no errors in the model. If the postulates are accepted, then Bell's theory is wrong. ... RE: Joy Christian's LHV Model that disproves Bell - gill1109 - 06-19-2016
(06-18-2016, 07:31 PM)FrediFizzx Wrote:Please take a careful look at equations (1) and (2) of Christian's revised one-page paper http://arxiv.org/pdf/1103.1879v2.pdf . The functions A and B, as they are defined on the left hand sides of these equations, certainly do take values in \(S^3\). But these values can be explicitly computed if one knows the definitions of D and L and some elementary results of geometric algebra. One finds(06-18-2016, 02:59 PM)gill1109 Wrote:... \(A(a,\lambda)=+1\) if \(\lambda = +1\), \(A(a,\lambda)=-1\) if \(\lambda = -1\); \(B(b,\lambda)=-1\) if \(\lambda = +1\), \(B(b,\lambda)=+1\) if \(\lambda = -1\). As Christian himself writes explicitly on the right hand sides of (1) and (2). So according to definition (4) the correlation is "-1". Hence there must be mistakes in Christian's calculations (5) to (12). I am accepting Christian's postulates. For the definitions of D and L Christian refers to http://arxiv.org/pdf/1501.03393v6.pdf, see in particular equations (5) and (9) in that paper. We find D(n) = I n, L(n, lambda) = lambda I n. One needs to know the standard multiplication rules of geometric algebra; in particular, the pseudoscalar I commutes with everything and its square is -1. |