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Richard,
You’re jumping to conclusions without reading my papers nor understanding what I have done.
I do not posit “two spins.” The model is one spin-1 bivector
with two blades; spin is defined as
and its scalar/wedge parts
are givien in Section 2.5, Eqs. 11–12. Please point to an error there.
Sanctuary, B. The Classical Origin of Spin: Vectors Versus Bivectors. Axioms 2025, 14, 668. https://www.mdpi.com/2075-1680/14/9/668/pdf
No exotic apparatus required. Standard Stern–Gerlach–type settings probe projections; the extra correlation arises because EPR experiments measure correlations, to which both vector (polarization) and bivector (coherence) parts contribute.
“Correlations can’t be added” is exactly backwards here. You think of the lever law in a single classical convex set, and there I agree. The paper shows the full EPR curve (cosine-like) is the sum of a vector term (CHSH≈2) and a bivector term (CHSH≈1); they exist in complementary convex sets, so Bell’s inequalities apply to each separately. Please see Section 3. Correspondence and Parity, and disagree or not: note that you cannot say it is out of your background. The argument is standard math of classical convex sets, and is not challenging .
If you think my decomposition is equivalent to your “A-spin/B-spin” toy, please indicate which step in Eqs. 11–12 (or the immediate quaternion form) makes that identification valid, or show the algebraic inconsistency.
So you do not need to start a new HV theory, I have already done it without HV, using classical mechanics, and rejecting the SM with a clear ontic and local description of a bivector which does one thing: changes the algebra from Cl(1,3) to Cl(2,2), see the figure. So you have not in any way challenged my hypothesis that the SM be replaced by the BiSM.
Bryan


Richard said1. "The curve can be written as a sum of two curves. That can be done in a myriad of ways, it proves nothing."Bryans reply:The curve is decomposed according to the geometric product:
<image.png>Go ahead, use any other math decomposition, and you lose the separation into symmetric (polarization) and antisymmetric (coherence).2. "So according to you, is each pair of binary outcomes the sum of two pairs of binary outcomes?"Bryan's reply:No, I say that each binary outcome is correlation between either two symmetric vectors or between two antisymmetric bivectors. Better detectors are always desirable.3. "If so, please get an experimentalist to do an enhanced Bell experiment with two binary outcomes at each measurement station for each “particle pair”.Bryan's reply:It is clear you have not read nor are interested in my paper, https://www.mdpi.com/2075-1680/14/9/668/pdf In section 8, I suggest 6 new or modified experiments. 8.2 amd 8.3 address your suggestion to separate coincidences from symmetric and antisymmetric. Yes, I hope that experimentalists take up that challenge.4. You also, everywhere, state I must average those two outcomes not add. I have fully addressed that in the paper, (convex sets section 3) but if we follow your prescription of averaging, then we must average the four coincidences from CHSH and not add them. This, then, is Richards CHSH equation, averaged.
<image.png>
Both Richard and I have responded to your critique.
We have studied your work in detail.
Your work is fundamentally flawed.
Your version of the geometric product is a complete misunderstanding of the concepts.
Both Richard and I have patiently attempted to explain this to you over and over.
It does not matter what we say because you are not willing to listen.
Your claims are complete confused nonsense that will disappear without a trace.
I will again quote Clifford Truesdell "This paper gives wrong solutions to trivial problems. The basic error, however, is not new."
/Jan-Åke
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