Joy Christian's LHV Model that disproves Bell FrediFizzx Independent Researcher Posts: 85 Threads: 1 Joined: Jun 2016 Reputation: 0 06-14-2016, 07:09 AM (06-14-2016, 06:17 AM)Schmelzer Wrote: (06-13-2016, 07:44 PM)FrediFizzx Wrote: Let me be more mathematically specific.  It really is quite simple.  You do the correlation calculation in $$S^3$$ and get the result of $$-a \cdot b$$.  Now after backtracking, any set of +/-1 outcomes that satisfy that result will do.  Ya take that set and do the correlation calculation in $$R^3$$ like they do for the experiments.  You will find that the result is $$-a \cdot b$$. So simple.What "backtracking"?   Or you have well-defined functions $$A(a,\lambda), B(b,\lambda)$$ with values in $$\{-1,1\}$$, together with a probability distribution $$\rho(\lambda)d\lambda$$, or you don't.... What it boils down to is that you are rejecting the $$S^3$$ postulate if you don't want to allow the correlation calculation to be done in a $$S^3$$ manner.  Anyways, I have wasted enough time on this.  We have got exciting new physics to work on! ... gill1109 Junior Member Posts: 46 Threads: 0 Joined: Jun 2016 Reputation: 0 06-14-2016, 08:04 AM (This post was last modified: 06-14-2016, 08:08 AM by gill1109.) (06-13-2016, 07:44 PM)FrediFizzx Wrote: (06-13-2016, 07:31 PM)Schmelzer Wrote: (06-13-2016, 07:14 PM)FrediFizzx Wrote: ???  I just told you how to do it if you want to do the correlation calculation in $$R^3$$.I'm not guided by wishful thinking, so it does not matter for me what one has to do if one wants to reach some result one likes.  I care about what is possible if  $$A(a,\lambda), B(b,\lambda)\in \{-1,1\}$$, and in this case I have a simple, elementary proof which shows me that Bell's inequality will be the result. The only reason you think it is "wishful thinking" is because your prejudice is so high it is blocking you from seeing the truth.  Let me be more mathematically specific.  It really is quite simple.  You do the correlation calculation in $$S^3$$ and get the result of $$-a \cdot b$$.  Now after backtracking, any set of +/-1 outcomes that satisfy that result will do.  Ya take that set and do the correlation calculation in $$R^3$$ like they do for the experiments.  You will find that the result is $$-a \cdot b$$. So simple. ...You can do the theoretical correlation calculation in $$S^3$$ (the set of unit quaternions). If you do it right you will find that the result is -1. The model is:  $$A(a,\lambda) = \lambda, B(b,\lambda)= - \lambda, \lambda \in \{-1,1\}$$. The outcomes -1 and +1 are to be thought of as elements of the set of unit length quaternions $$S^3$$. Correlation is defined in the usual way. It is easy to find the mistake in Christian's computation (5)-(9). Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 06-14-2016, 08:14 AM (06-14-2016, 07:09 AM)FrediFizzx Wrote: ... What it boils down to is that you are rejecting the $$S^3$$ postulate if you don't want to allow the correlation calculation to be done in a $$S^3$$ manner.  Anyways, I have wasted enough time on this.  We have got exciting new physics to work on! ...What I'm rejecting is that any proposals for using some $$A(a,\lambda), B(b,\lambda) \in S^3$$ have anything to do with Bell's theorem, except in the trivial case where $$A(a,\lambda), B(b,\lambda) \in \mathbb{Z}_2 \subset S^3$$, and in this case embedding it into $$S^3$$ is as meaningless as embedding it into the Monster group. Fine if you have some new physics, and if this stops you loosing time "disproving" simple theorems. FrediFizzx Independent Researcher Posts: 85 Threads: 1 Joined: Jun 2016 Reputation: 0 06-14-2016, 05:44 PM (This post was last modified: 06-14-2016, 08:06 PM by FrediFizzx.) (06-14-2016, 08:14 AM)Schmelzer Wrote: (06-14-2016, 07:09 AM)FrediFizzx Wrote: ... What it boils down to is that you are rejecting the $$S^3$$ postulate if you don't want to allow the correlation calculation to be done in a $$S^3$$ manner.  Anyways, I have wasted enough time on this.  We have got exciting new physics to work on! ...What I'm rejecting is that any proposals for using some $$A(a,\lambda), B(b,\lambda) \in S^3$$ have anything to do with Bell's theorem, except in the trivial case where $$A(a,\lambda), B(b,\lambda) \in \mathbb{Z}_2 \subset S^3$$, and in this case embedding it into $$S^3$$ is as meaningless as embedding it into the Monster group.  ... That's right, stay stuck in "flatland" if you wish.  What you fail to realize is that Bell's theory is "rigged" to be stuck in $$R^3$$.  You can't get out if you rigidly follow Bell's prescription.  Bell's theory is junk physics allowing nothing new.  His model prescription for LHV models is wrong.  Time for something new. ... Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 06-15-2016, 05:34 AM Rigged or not, it falsifies a whole class of theories. This is, indeed, rigged to a particular experiment, the one which has $$\mathbb{Z}_2$$ results. Feel free to speculate about fairies or other things which are not rigged to particular experiments here on Earth. FrediFizzx Independent Researcher Posts: 85 Threads: 1 Joined: Jun 2016 Reputation: 0 06-15-2016, 07:28 AM (This post was last modified: 06-15-2016, 07:32 AM by FrediFizzx.) (06-15-2016, 05:34 AM)Schmelzer Wrote: Rigged or not, it falsifies a whole class of theories.  This is, indeed, rigged to a particular experiment, the one which has $$\mathbb{Z}_2$$ results.   Feel free to speculate about fairies or other things which are not rigged to particular experiments here on Earth.... If $$\mathbb{Z}_2$$ results is your criteria then Bell's prescription also eliminates quantum mechanics.  That makes his theory wrong.  Go ahead and produce the negative cosine curve for the EPR-Bohm scenario with QM using +/- 1 outcomes.   However, we believe the QM prediction is correct and Joy Christian's model actually explains it.  You are left with a non-local HV model that doesn't really explain anything as far as quantum mechanics is concerned.  Do you even have a non-local HV model? Anyways, I think I prefer a classical local-realistic explanation for quantum correlations. I think others will prefer that also. ... Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 06-15-2016, 10:24 AM There is no problem nor for dBB theory, nor for QM. You may have heard that dBB theory is equivalent to standard QM in its predictions, and it is clearly and obviously realistic. It is not Einstein-causal, QM is not Einstein-causal too, so that Bell's theorem cannot be proven for above theories. The whole point of Bell's theorem was, for Bell, to prove that the accusation against dBB theory that it is not Einstein-causal is nonsensical - no realistic interpretation can be Einstein-causal. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 06-16-2016, 12:12 AM Hello FreddiFizzx, I would appreciate it if you'd explain Joy Christian's approach as follows. Nick Herbert, in his 1985 book, provided a simple way to see the point of Bell's inequality. I'd like to describe it, then you explain how it works in Christian' approach. Suppose that Alice and Bob both set their detectors at 0 degrees. We get perfect correlation. Since spins must be opposite that means in every detection event one of them gets -1, the other 1. Now Alice sets detector to -30 degrees. QM predicts correlation of cos^2, namely 3/4. Thus 1 out of 4 are "wrong", meaning, in this case, equal to each other (since they should be opposite). If Alice sets detector back to 0 degrees, and Bob to 30 degrees, again 1 out of 4 are wrong. Now, Alice sets detector to -30 degrees. The total separation is now 60 degrees. cos^2 is only 1/4. So QM predicts 3 out of 4 are wrong. (On average, of course). This is a logical contradiction. Note, of course the 0 setting doesn't exist when they're at +- 30 degrees; but imagine it does exist. Compared to the 0 setting, they each should have only 1 out of 4 wrong. If they get the same one wrong, those two "wrongs" cancel each other and we get a correlation. So the worst case is when none of their misses match. That gives exactly 2 out of 4, 1/2, correlated, and 1/2 missing - maximum. But in the actual run - forgetting about the imaginary 0 setting - QM says they only get 1/4 correlated, 3/4 wrong. It's impossible to reproduce this result with local hidden variables, as far as I know. Now, you say Christian's approach does manage to reproduce these same QM statistics with a LHV theory. It involves using a different topological space and other complications. Please discuss this simple thought experiment and show how Christian's approach explains it. Thanks. FrediFizzx Independent Researcher Posts: 85 Threads: 1 Joined: Jun 2016 Reputation: 0 06-16-2016, 01:08 AM (06-16-2016, 12:12 AM)secur Wrote: Please discuss this simple thought experiment and show how Christian's approach explains it. Thanks. Hi secur, It is a bit complicated so probably best for you to read this paper by Joy Christian where he explains it. http://arxiv.org/abs/1405.2355 "Local Causality in a Friedmann-Robertson-Walker Spacetime" Feel free to ask any questions about the content in the paper and I will try to answer them.  I suspect Ilja is too prejudiced to read it but perhaps you will and we can discuss it here. ... secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 06-16-2016, 01:48 AM (This post was last modified: 06-16-2016, 01:49 AM by secur.) It seems that he's representing spin states as quaternions; not sure of the details. The Pauli matrices which we usually use to represent spin are, of course, equivalent algebraically to quaternions. Dirac, BTW, originally used quaternions to derive his relativistic spin equation. Anyway quaternions give the same correlations as QM, since they can be used in place of Pauli matrices. Christian alludes to this correlation, cos^2 + sin^2, in equation 19. So it appears that in effect he's just restoring Dirac's original quaternionic representation for spin. In that case, the paper essentially (at best) amounts to a new ontological interpretation. Explains why QM spin works as it does: because we live in a FLRW world. This is just off the top of my head, I may be thoroughly misunderstanding. What do you think? « Next Oldest | Next Newest »