Sci.Physics.Foundations

In this paper, Karl Hess clearly debunks the Bell-CHSH myth. It is the clearest presentation of the problem I've seen so far. https://www.scirp.org/journal/paperinformation.aspx?paperid=126265#ref8 The description from Section 2 all the way to Section 3.1 must be earth-shattering for quantum mysteri...

It looks very interesting. I was wondering why they used quaternions instead of geometric algebra. Perhaps their initial interest is to show equivalence to QM. But I think they will make their work clearer by using geometric algebra directly.

Well Ok, thanks for trying but you seem to think a concise definition for "hidden variable" is not required any more. It can just be a free for all and defined however one wants to define it. To me, that is a bunch of nonsense. But here is what Joy and I agreed to for the definition, 1. A...

Well, thanks but it doesn't resolve the dispute. Joy claims the singlet spin vector, "s", above is a hidden variable. I say it is not a hidden variable since quantum mechanics knows about it via the singlet wavefunction. What say you? What we have here is two different local EPR-Bohm mode...

The Original EPR paper proved that the Quantum Mechanical Wavefunction was not a complete description of physical reality: Previously we proved that either (1) the quantum-mechanical description of reality given by the wave function is not complete or (2) when the operators corresponding to two phys...

Fred are you saying, your functions above do not satisfy Bell's equation (1) of. A(a,h) = -B(a,h) = +/-1? If not, then what makes you think they have any relationship to Bell's theorem? If they do, then Bell's equation (2) follows also. I don't understand how you can claim Bell's equation (2) is zer...

Fred is right! Can someone please explain to me why Bell's equation 2 is equal to zero. What am I missing? It is mainly because Bell didn't specify any actual functions for A and B other than equal to +/-1. . Fred, if this is what you and Richard are talking about, then you are very wrong. A(a,h) i...

Michel wants to deny any metaphysical assumption behind the claim p(h|a, b) =/= p(h). I can't really engage with a misrepresentation of my position which I've explained multiple times. p(h|a, b) =/= p(h) has absolutely no metaphysical "assumption" behind it. It's simply a statement that t...

gill1109 wrote: ↑Mon Dec 27, 2021 11:34 pm Fred is right!

Can someone please explain to me why Bell's equation 2 is equal to zero. What am I missing?

... This is what is measured experimentally when we measure E(a,b). No it is NOT! It is equal to zero and they don't get zero in the experiments. Use a better equation or stop talking about it. . Then I don't know what you are talking about. I don't see how you get zero. Even if you reduce it to E(...

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification. . p(h|ab) is the distribution that corresponds to the outcomes measured at settings (a,b). p(h|ab) is not the source distribution. Why do you insist on using the source distr...

We have stopped communicating. You have been ignoring the points I have been making, so I will do the same. We are not going to agree about this. Not true. I'm making a good-faith effort to address your points. This is not about agreement. It is possible to state an argument accurately even if you ...

You are making the assumption that the probability distribution p(h) depends on the settings a and b. You have to justify that assumption. Superdeterminism means \rho(\lambda) \neq \rho(\lambda|ab) This absolutely does not mean \rho(\lambda) depends on the settings (a, b). It simply means the distr...

E(c, d) = \int_H A(c, h) B(d, h) \; p(h)\; dh will be wrong, So your expression is the stronger position and demands a physical justification. What is it? When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification. . p(h|ab) is the di...

I am talking about E(a, b) = \int_H A(a, h) B(b, h) \; p(h)\; dh , with the following physical justification: So what does H and p(h) represent in the bomb explosion example where "h" is the momentum of the fragment and A(a,h) is the measurement function representing my machine? Are you s...

What is the physical justification for that assumption? This is the crux of the matter. The physical justification for using p(h) as the probability distribution is quite simple. It is the distribution of microstates or hidden variables that are being integrated over at a later time. In experiments...

Restricting the domain of A(a, h) in that way prevents A(a, h) from being well defined for all h in H, which is the space of all hidden variables for the physical system. So you chose to assume that A(a,h) must be defined for all values of "h". Why? What physical justification do you have...

The main issue is that the setting-dependent distribution function p(a, b | h) does not make much physical sense. I don't know what physical role you think p(h|a, b) plays that you just can't accept? Whether you like it or not, p(h|a, b) ALWAYS plays a physical role in the EPRB experiments. The onl...

gill1109 wrote: ↑Sun Dec 26, 2021 11:44 pm You keep making an assumption of bad faith on my part.

My assumption is reasonable based on the history concerning this topic:
http://www.sciphysicsforums.com/spfbb1/ ... ple#p14472

Joy Christian wrote: ↑Mon Dec 27, 2021 6:55 am Indeed. So Michel's function A(a, h) = 1/sign(a.h) is not as well defined as he thinks it is. It blows up to infinity for a.h = 0.

Yeah, if you go outside the domain of the function, what do you expect? Every function is not "well-behaved" outside its domain. Garbage in, garbage out.