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Bell's theorem - for or against Hidden Variables?
secur wrote:  "Anyway, we'll have to 'agree to disagree' about the fixed frame. IMHO my comments are correct."

That's collegial and fair.  All I can ask is that you please consider my argument, as carefully as I have considered yours.

It is of logical necessity that fixed frames cannot exist in a four-dimension, connected and continuous, spacetime.  This is a property among others that led Einstein to the coordinate-free geometry of general relativity.  What is true in GR cannot be false in SR.

secur: "Re. 'all physics is local': like most science, this can never actually be proven, only disproven."

Falsification is indeed the highest honor that one can bestow on a scientific theory.  In this case, though, we are not speaking of a theory; rather, a measurement framework.  As Wheeler said, "No phenomenon is a real phenomenon until it is an observed phenomenon."  

And that is what Joy Christian promises:  a coordinate-free framework.  How does one have a coordinate-free framework with a privileged observer?  Joy adds the concept of nature's random choice, that stands in for the role of time -- a nonlinear time that forces the framework to be analytical.  An analytical measure is dependent on boundary conditions.  

So the boundary conditions are built into the 3-dimension ball analyzed in 4-dimension spacetime.  The parts share a rest frame initially.  http://www.relativitybook.com/resources/...metry.html Einstein, from "Geometry and Experience":  "All practical geometry is based upon a principle which is accessible to experience, and which we will now try to realise. We will call that which is enclosed between two boundaries, marked upon a practically-rigid body, a tract. We imagine two practically-rigid bodies, each with a tract marked out on it. These two tracts are said to be 'equal to one another' if the boundaries of the one tract can be brought to coincide permanently with the boundaries of the other. We now assume that:

If two tracts are found to be equal once and anywhere, they are equal always and everywhere."

Can't do that with dice.

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RE: Bell's theorem - for or against Hidden Variables? - by Thomas Ray - 08-31-2016, 09:47 PM

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