Login Register

Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Bell's theorem - for or against Hidden Variables?
(06-07-2016, 04:59 AM)Schmelzer Wrote:
(06-07-2016, 12:00 AM)FrediFizzx Wrote:
(06-06-2016, 08:27 PM)Schmelzer Wrote: The linked paper shows only that Joy Christian has not understood why Bell can do this.  It is an application of the EPR argument.

Nonsense.  Please demonstrate how ⟨Ak(a)Bk(b)+Ak(a)Bk(b′)+Ak(a′)Bk(b)−Ak(a′)Bk(b′)⟩ could represent something that is actually physical.

If Alice measures in direction a, and Bob too, then they get the 100% correlated result.  But Alice decision to measure a does in no way disturb Bob's measurement - which is the Einstein causality assumption.  So, following the EPR criterion of reality, the value measured by Bob has to be well-defined by the initial state already before the measurement is done.  This predefined result is \(A(a,\lambda)\).  

Once these functions objectively have to exist, and predefine the measurement results, we can use them to compute the \(E(a,b)\) as being 
\[E(a,b) = \int A(a,\lambda)B(b,\lambda)\rho(\lambda)d\lambda\] and then we can measure them.   As usual in statistic experiments, for a small number the result may differ a lot, but in the long run, for a large enough number, you will get the observed \(E(a,b)\) quite close to the one predicted by the theory, if the theory is correct.  

But once these functions objectively have to exist, you can compute in theory even sums which you cannot measure.  Because the theory itself has to be mathematically consistent.

So, the whole expression is not measured in a single experiment.  But different experiments are used to define the different parts.  It is a theoretical consideration which shows that in an Einstein-causal theory this is possible and gives the same results.

Yes, that is the standard "party line" on how to try to justify flawed mathematics and flawed physical reasoning. If you want to continue believing that, then there is probably nothing I can do to convince you otherwise. But you did highlight the exact flaw, "So, the whole expression is not measured in a single experiment." So excuse me if your argument is not at all convincing physically. I think we are finished here. Lurkers can decide for themselves.

Messages In This Thread
RE: Bell's theorem - for or against Hidden Variables? - by FrediFizzx - 06-07-2016, 06:40 AM

Forum Jump:

Users browsing this thread: 20 Guest(s)