Comments on: Bell’s-inequality-denialist Joy Christian offers me $200K if scalable quantum computers are built

By: Tsirelson’s bound | MyCQstate

Tsirelson’s bound | MyCQstate — Wed, 10 Oct 2012 18:56:19 +0000

[...] of the EPR paradox and Bell’s work were still being debated (well, they still are even today…). As we saw in the previous post, 20 years earlier Bell had demonstrated that the set of [...]

By: Elevated Doofus Activity « Pink Iguana

Elevated Doofus Activity « Pink Iguana — Wed, 16 May 2012 10:29:08 +0000

[...] I was Wrong about Joy Christian, here. Sometimes the snark has go to eleven. In response to my post criticizing his “disproof” of Bell’s Theorem, Joy Christian taunted me that “all I knew [...]

By: James Putnam

James Putnam — Tue, 15 May 2012 21:22:44 +0000

My last message was not properly punctuated:

“…that explains quantum theory.”

What about that “…tricky continuous measurement function…” Anyone is invited to help with this.

James

By: Thomas H Ray

Thomas H Ray — Tue, 15 May 2012 20:59:05 +0000

Richard, glad to see you admit that you’ve run completely out of argument. Thanks.

By: Joy Christian

Joy Christian — Tue, 15 May 2012 20:58:12 +0000

Sandro,

I agree up to your point 2).

Your point 3) is not quite right. Let me rephrase it in my words:

3) We have functions w(a)w(a, L) and w(b)w(b, L) for which we can write A(a, L) = w(a)w(a, L) and B(b, L) = w(b)w(b, L). These functions are [ my ] cases definitions. What is important is that these equalities hold for the whole domain of the variables L, a, and b. Clearly to state this equality, the codomains of A and w(a)w(a, L), as well as those of B and w(b)w(b, L), have to be the same.

They are the same. It is the parallelized 3-sphere, S^3.

By: James Putnam

James Putnam — Tue, 15 May 2012 20:54:46 +0000

Joy and David Brown,

David Brown Says:
Comment #613 May 15th, 2012 at 1:23 pm

“From the viewpoint of Christian’s theory of local realism, the universe’s topology of allowable measurements forbids Alice from making a measurement that would provide a counterexample to Bell’s theorem. Alice does not need to whirl herself around the universe because the problem does not arise — there is a tricky continuous measurement function that explains quantum theory. Is David Brown wrong here?”

I found David’s remark compelling and waited for an authoritative response.

Joy: “David Brown is more right than most wrongs on this blog.”

I need more to be said. “… Alice does not need to whirl herself around the universe because the problem does not arise — there is a tricky continuous measurement function that explains quantum theory. What about that “…tricky continuous measurement function…” Anyone is invited to help with this.

James

By: Joy Christian

Joy Christian — Tue, 15 May 2012 20:37:36 +0000

David Brown asks: “Is David Brown wrong here?”

David Brown is more right than most wrongs on this blog.

By: Richard Gill

Richard Gill — Tue, 15 May 2012 19:48:17 +0000

Alex V #609. Eq. (4) defines two different bases of the same algebra. Different multiplication tables, same algebra. Nothing to do with Alice vs. Bob.

Tom #612. You are talking about your own private fantasy, not about Joy’s model. You talk about your imagination, your heartfelt desire and deep belief, that this would be what it’s about. Unfortunately your poetic science-fancy is even further dissociated from the actual maths which Joy writes down, than his own sales patter is.

By: Sandro

Sandro — Tue, 15 May 2012 18:37:53 +0000

Too bad, same typo as before (should avoid copy and paste). B(a,L)=g(L) should read B(b,L)=g(L).

By: Sandro

Sandro — Tue, 15 May 2012 18:35:30 +0000

Joy# 610: Dear Joy, as you very correctly remarked, I could not make an articulate sentence in the language of geometric algebra, and I won’t even try to, don’t worry. I wouldn’t say that mine was a confession, as I never stated I am an expert on the topic, or tried to hide the fact that I’m not. But let’s not diverge. All I want is to understand what you mean on a rather basic notational level, and hence I’ll let your comments guide the discussion.

So, so far it seems like we concluded that:

1) We have variables L, a, and b, each living in some fixed domain, as by you specified.

a) L, a, and b are mutually independent. That is, neither a or b depend in any way on L, neither L depends on a and b.

If you agree, as you seem to do, we can move on points 2 and 3, which were the following.

2) We have functions A(a,L) and B(b,L). As far as I’m concerned, you can define their codomains as you whish. Their domain is clearly the cartesian products of the domains of the corresponding variables.

3) We have functions f(L) and g(L) for which we can write A(a,L)=f(L) and B(a,L)=g(L). These functions are your cases definitions. What is important is that these equalities hold for the whole domain of the variables L, a, and b. Clearly to state this equality, the codomains of A and f, as well as those of B and g, have to be the same.

Can you comment?