Eve fools Alice and Bob

[FrediFizzx]

http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404

Paper is here,

http://arxiv.org/abs/1106.3224
"Experimentally faking the violation of Bell's inequalities"

Evidentially these folks are not aware of Joy Christian's work.

http://arxiv.org/find/grp_physics/1/au:+christian_joy/0/1/0/all/0/1
http://www.fqxi.org/community/forum/topic/995
"On the Origins of Quantum Correlations"

Best,

Fred Diether

[mpc755]

On Dec 15, 2:40 am, "FrediFizzx" <fredifi...@hotmail.com> wrote:
> http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404
>
> Paper is here,
>
> http://arxiv.org/abs/1106.3224
> "Experimentally faking the violation of Bell's inequalities"
>
> Evidentially these folks are not aware of Joy Christian's work.
>

> http://arxiv.org/find/grp_physics/1/au:+christian_joy/0/1/0/all/0/1http://www.fqxi.org/community/forum/topic/995

> "On the Origins of Quantum Correlations"
>
> Best,
>
> Fred Diether

The concept behind Bell's Theorem is fundamentally flawed. There is no
experiment which has a classical and a non-classical predicted result.

Pilot-wave theory is the classical interpretation of quantum mechanics
where the particle has a well defined trajectory which takes it
through one slit while the associated physical wave passes through
both. Pilot-wave theory uses the same mathematics as all other
interpretations of quantum mechanics and therefore makes the same
predictions.

[glen herrmannsfeldt]

mpc755 <mpc...@gmail.com> wrote:
> On Dec 15, 2:40 am, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>> http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404

(snip)

> The concept behind Bell's Theorem is fundamentally flawed. There is no
> experiment which has a classical and a non-classical predicted result.

Reminds me again of something I saw some time ago, but then can't find
a reference for. Something like the biggest pseudo-scientific idea
foisted on our students as science is (and creationism immediately
comes to mind) the Copenhagen interpretation of Quantum Mechanics.

If anyone happens to have a reference, I would like to know about it.
I tried the usual search engines but, it seems, without the right query.

> Pilot-wave theory is the classical interpretation of quantum mechanics
> where the particle has a well defined trajectory which takes it
> through one slit while the associated physical wave passes through
> both. Pilot-wave theory uses the same mathematics as all other
> interpretations of quantum mechanics and therefore makes the same
> predictions.

-- glen

[FrediFizzx]

"glen herrmannsfeldt" <g...@ugcs.caltech.edu> wrote in message
news:jcdhrp$bqn$1...@speranza.aioe.org...

> mpc755 <mpc...@gmail.com> wrote:
>> On Dec 15, 2:40 am, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>>> http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404
>
> (snip)
>> The concept behind Bell's Theorem is fundamentally flawed. There is no
>> experiment which has a classical and a non-classical predicted result.
>
> Reminds me again of something I saw some time ago, but then can't find
> a reference for. Something like the biggest pseudo-scientific idea
> foisted on our students as science is (and creationism immediately
> comes to mind) the Copenhagen interpretation of Quantum Mechanics.
>
> If anyone happens to have a reference, I would like to know about it.
> I tried the usual search engines but, it seems, without the right query.

Perhaps it was Jaynes that said something like that? Doesn't really matter
now since Joy Christian has most likely solved the mystery about quantum
correlations.

>> Pilot-wave theory is the classical interpretation of quantum mechanics
>> where the particle has a well defined trajectory which takes it
>> through one slit while the associated physical wave passes through
>> both. Pilot-wave theory uses the same mathematics as all other
>> interpretations of quantum mechanics and therefore makes the same
>> predictions.

Unfortunately, pilot-wave theory is still non-local in an EPRB type
scenario.

Best,

Fred Diether

[Daryl McCullough]

On Thursday, December 15, 2011 1:18:44 PM UTC-5, mpc755 wrote:

> The concept behind Bell's Theorem is fundamentally flawed. There is no
> experiment which has a classical and a non-classical predicted result.

No, it is not fundamentally flawed. It is a theorem
about the limits to what kind of correlations between
random variables are possible without causal interactions.
Neither Joy Christian's work nor that of De Raedt are
disproofs of Bell's theorem, and it is misleading to
call them that. You don't disprove a theorem except by
discovering a mathematical error in it. At most, they
could show that the theorem does not apply to real
experiments, for various reasons.

[Daryl McCullough]

On Thursday, December 15, 2011 2:40:31 AM UTC-5, FrediFizzx wrote:
> http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404
>
> Paper is here,
>
> http://arxiv.org/abs/1106.3224
> "Experimentally faking the violation of Bell's inequalities"
>
> Evidentially these folks are not aware of Joy Christian's work.

Joy Christian's work is *NOT*, despite his article titles, a
disproof of Bell's Theorem. That isn't to say that there are
mistakes in his mathematical derivations, but that his conclusions
are not relevant to Bell's claims.

Let me simplify the spin-1/2, twin-particle version of the EPR
thought experiment to illustrate.

A deterministic hidden-variables explanation of the EPR would
be a set V of values for a hidden-variable lambda, together
with a pair of functions f(v,u) and g(v,u) taking a parameter
v from V and a unit vector u, and returning +1 or -1, where f
and g satisfied certain statistical properties. Bell essentially
proved that there are no such functions. Joy Christian certainly
did not prove otherwise. He changed the problem statement to
one that had a solution. There is nothing wrong about doing
that, but it is *COMPLETELY* incorrect to call it a disproof
of Bell's theorem.

[Daryl McCullough]

On Thursday, December 15, 2011 4:06:05 PM UTC-5, glen herrmannsfeldt wrote:

> Reminds me again of something I saw some time ago, but then can't find
> a reference for. Something like the biggest pseudo-scientific idea
> foisted on our students as science is (and creationism immediately
> comes to mind) the Copenhagen interpretation of Quantum Mechanics.

What annoys me is that for the most intractable problems in science,
including the interpretation of quantum mechanics and a quantum theory
of gravity, everyone has his own pet solution and considers every other
approach to be pseudo-scientific nonsense.

The advantage to the Copenhagen interpretation is that it allows
scientists to get on with science without waiting for the
definitive answer to the philosophical question of the meaning of
quantum mechanics. Copenhagen is a *pragmatic* approach to using
quantum mechanics, which treats macroscopic phenomena as essentially
classical, while treating microscopic phenomena quantum mechanically.
It really is not a scientific theory, and doesn't pretend to be,
but is a rule of thumb for using quantum mechanics.

[mpc755]

Yes, pilot-wave theory is non-local and that is what is incorrect in
pilot-wave theory.

When a downconverted photon pair are created the original photons
momentum is conserved and the downconverted photon pair are created
with opposite angular momentums and will be detected with opposite
spins as predicted by all interpretations of quantum mechanics.

There is no reason for pilot-wave theory to be non-local when you
understand conservation of momentum applies to the downconverted
photon pair meaning the photons are created with opposite angular
momentums.

[Joy Christian]

On Dec 16, 7:11 pm, Daryl McCullough <stevendaryl3...@yahoo.com>
wrote:

>
> Joy Christian's work is *NOT*, despite his article titles, a
> disproof of Bell's Theorem. That isn't to say that there are
> mistakes in his mathematical derivations, but that his conclusions
> are not relevant to Bell's claims.
>
> Let me simplify the spin-1/2, twin-particle version of the EPR
> thought experiment to illustrate.
>
> A deterministic hidden-variables explanation of the EPR would
> be a set V of values for a hidden-variable lambda, together
> with a pair of functions f(v,u) and g(v,u) taking a parameter
> v from V and a unit vector u, and returning +1 or -1, where f
> and g satisfied certain statistical properties. Bell essentially
> proved that there are no such functions. Joy Christian certainly
> did not prove otherwise. He changed the problem statement to
> one that had a solution. There is nothing wrong about doing
> that, but it is *COMPLETELY* incorrect to call it a disproof
> of Bell's theorem.

I disagree. I have decisively disproved Bell’s theorem.

Please read this paper: http://arxiv.org/abs/1103.1879

Then please read at least the section IV of this paper:
http://arxiv.org/abs/1110.5876

Finally, read the introduction of the last paper.

It is time to recognize that Bell made a mistake in his
theorem. More importantly, it is time to move on and
recognize the true significance of quantum correlations.

Joy Christian

[robert bristow-johnson]

Daryl,

perhaps i should just email this to you, but i want to say that posts
like this and others ever since this one:

http://groups.google.com/group/sci.physics.research/msg/a0ec7e6c84fc8499

have garnered a great deal of respect from me. what you are to
sci.physics.[research|foundations] is what i hope to approximate at
comp.dsp .

sorry to gush, but you hit the nail so squarely on the head with
comments like this.

--

r b-j r...@audioimagination.com

"Imagination is more important than knowledge."

[mpc755]

On Dec 16, 2:11 pm, Daryl McCullough <stevendaryl3...@yahoo.com>
wrote:

There need not be any hidden variables in pilot-wave theory and it
need not be non-local.

All you need to do to understand the correctness of pilot-wave theory,
or what underlies pilot-wave theory where the physical particle has an
associated physical wave, is to understand the downconverted photon
pair are created with opposite angular momentums. Knowing this allows
for the predictions of all interpretations of quantum mechanics to
apply to pilot-wave theory and for the theory to not be a hidden
variable theory and for the theory to be local.

[ben6993]

On Dec 16, 7:11 pm, Daryl McCullough <stevendaryl3...@yahoo.com>
wrote:

Below is what I think I have learned from the Bell discussions on
earlier threads:

1. Joy Christian has apparently shown that, if you allow D>4
dimensions (D=7), the two entangled electrons are not genuinely
'entangled'. They always have opposite states, just as a black sock
is never a white sock. One might think that N=7 dimensions opens up
more scope for there to exist non-locality, but the abolition of
genuine 'entanglement', ie the abolition of simultaneous
superposition of different spin states for the one particle, is enough
to imply non-locality. I intend to learn enough Clifford Algebra to
understand the paper but am still following the Stanford online QM
course.

There is always a probability of .5 that an electron will be say 'up'
when measured. That is in the 4D laboratory space of real
measurements in which full access to the spin state information is not
available. If you take into account the full dimensions of the
electron, then it is in one state or the other, not in a
superposition. The superposition can therefore only be an effect of
the observer's uncertainty.

Feynman's path integral seems to be proof that an electron is in
superimposed states, but it is not clear to me that an electron is
simultaneously in different spin states when it is apparently
occupying different positions in the laboratory simultaneously. It
could be argued to be always in the same spin state in all its
different (at least in the forward moving in time) positions. It is a
superposition of a kind, but not necessarily one of different spin
states at the same time for the same particle.

2. Dr. Raedt's papers appear to show that the Aspect confirmation of
entangled particles breaking Bell's Inequality is flawed. The
techniques for timing of the arrival of bona fide entangled pairs and
thereby controlling which data are included in the results is,
unfortunately and coincidentally, apparently identical to a data
pruning exercise which cuts away data which does not break the Bell's
Equality, leaving behind data which does break the Bell's Inequality.
I am not sure if any experiment has truly shown that
Bell's Inequality is broken by entangled pairs?

3. Bell's Inequality itself is, of course, a simple finding using
elementary Venn Diagrams. If you use all the data in a Venn Diagram
then you cannot break the equality. If you start with a whole set of
data and then prune that data, you can apparently break the inequality
without resorting to using abstract vector space algebra, as Dr Raedt
et al have shown.

[dlzc]

Dear Daryl McCullough:

On Dec 16, 12:11 pm, Daryl McCullough <stevendaryl3...@yahoo.com>
wrote:

I believe this to be incorrect. If we were to find a frame in which
the vacuum speed of light was not c, would this not invalidate the
postulates of Relativity, and therefore disprove it? The mathematics
are merely steady logic applied to get from the hypothesis /
postulates, to a quantifiable result.

For example, Einstein made an initial estimate for the angular
distortion of light for Eddington to look for. He later corrected
this. Did Einstein disprove Relativity with the correction of a
mathematical error? What you say above seems to say so.

Or did I misunderstand?

David A. Smith

[glen herrmannsfeldt]

ben6993 <ben...@hotmail.com> wrote:

(snip)

> There is always a probability of .5 that an electron will be say 'up'
> when measured. That is in the 4D laboratory space of real
> measurements in which full access to the spin state information is not
> available. If you take into account the full dimensions of the
> electron, then it is in one state or the other, not in a
> superposition. The superposition can therefore only be an effect of
> the observer's uncertainty.

> Feynman's path integral seems to be proof that an electron is in
> superimposed states, but it is not clear to me that an electron is
> simultaneously in different spin states when it is apparently
> occupying different positions in the laboratory simultaneously.

There are some examples in Feynman Lectures on Physics, vol. III,
which might help. If you take a X polarized electron (or anything
with a magnetic moment), and put it through a Y-axis Stern-Gerlach
apparatus, it will split equally into two beams based on the
Y polarization. If you test the Y polarization on the beams, you find
one is up and one is down. If you test the X polarization of one
of the beams, you find 50% up and 50% down.

If instead of testing the polarization, you put them through an
inverted Stern-Gerlach apparatus, put the beams back together as one,
then it still has the X polarization.

> It could be argued to be always in the same spin state in all its
> different (at least in the forward moving in time) positions. It is a
> superposition of a kind, but not necessarily one of different spin
> states at the same time for the same particle.

The split beams have Y polarization, but haven't forgotten the X
polarization that they had previously. It isn't visible unless the
two beams are put back together.

-- glen

[FrediFizzx]

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:11942922.417.1324042389828.JavaMail.geo-discussion-forums@yqja5...

> On Thursday, December 15, 2011 2:40:31 AM UTC-5, FrediFizzx wrote:
>> http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.107.170404
>>
>> Paper is here,
>>
>> http://arxiv.org/abs/1106.3224
>> "Experimentally faking the violation of Bell's inequalities"
>>
>> Evidentially these folks are not aware of Joy Christian's work.
>
> Joy Christian's work is *NOT*, despite his article titles, a
> disproof of Bell's Theorem. That isn't to say that there are
> mistakes in his mathematical derivations, but that his conclusions
> are not relevant to Bell's claims.

They are exactly relevant since Bell's claims don't match physical reality
as Joy has shown.

> Let me simplify the spin-1/2, twin-particle version of the EPR
> thought experiment to illustrate.
>
> A deterministic hidden-variables explanation of the EPR would
> be a set V of values for a hidden-variable lambda, together
> with a pair of functions f(v,u) and g(v,u) taking a parameter
> v from V and a unit vector u, and returning +1 or -1, where f
> and g satisfied certain statistical properties. Bell essentially
> proved that there are no such functions. Joy Christian certainly
> did not prove otherwise. He changed the problem statement to
> one that had a solution. There is nothing wrong about doing
> that, but it is *COMPLETELY* incorrect to call it a disproof
> of Bell's theorem.

You obviously haven't studied Joy Christian's work sufficiently. But
yeah..................., we are pretty much way past all of what you
mention above and on to studying the origins of quantum correlations. That
is a very interesting part of his work. There is more to "spacetime" than
what has been previously thought.

Best,

Fred Diether

[ben6993]

On Dec 17, 12:14 am, glen herrmannsfeldt <g...@ugcs.caltech.edu>
wrote:

Thanks Glen. I obtained Feynman Lectures on Physics on your
recommendation a few months back and have been steadily working
through Vol III. Up to page 7.1 at the moment. You did mention this
kind problem before but I wasn't ready for it then, but will try it
now. I think that I am clear about the problem and it is demonstated
in Figs 6.4 and 6.5 of Vol III (2010 version). Assuming this is the
problem you mean? Feynman said one can skip chapter 6 although I did
read it, but understood why he was reasuring readers who did not
understand it straight away. Me too.

I reserve the right to fail to explain, in commonsense 3D terms,
Figures 6.5a and 6.5b using arrows/helices, as a full explanation of
hidden variables apparently requires 7D. Especially as Feynman cites
a complex variable phase change occurring when the Stern-Gerlach
apparatus is rotated. However, I think I can explain these two
figures despite that. They seem straightforward to me so I probably
do not fully understand them. I intend to post my detailed
explanation for you to see on the 'implications..' thread so as not to
clog this 'Eve fools Alice' thread.

Note to Fred: when is there likely to be something to read on the
follow up work "studying the origins of quantum correlations"?

Ben6993
Not a physicist

[harald]

"Daryl McCullough" <stevend...@yahoo.com> wrote in message

news:27069963.147.1324041602008.JavaMail.geo-discussion-forums@vbbhx10...

> On Thursday, December 15, 2011 1:18:44 PM UTC-5, mpc755 wrote:
> You don't disprove a theorem except by
> discovering a mathematical error in it.

That is true for a mathematical theorem. However, Bell's is a theorem of
physics: it's a claim about physical theories, based on physical
assumptions. Already the term "local realistic theory" does not belong to
mathematics.
Compare: http://plato.stanford.edu/entries/bell-theorem/

[Daryl McCullough]

As I said, I don't think that the word "disproof" can be meaningfully
applied to theorems. You can attack Bell's theorem on several grounds:
you can find a mistake in the derivation, or you can find a hidden
assumption that should be made explicit, or you can argue that assumptions
(explicit or implicit) don't apply in cases of interest, or you can argue
that Bell's formalization of the problem (the way he tries to capture
physical concepts such as "locality" and "realism") is not correct.

The terminology, "disproving Bell's Theorem" is not very
descriptive of any of these lines of attack.

[Daryl McCullough]

On Friday, December 16, 2011 5:16:36 PM UTC-5, dlzc wrote:

> I believe this to be incorrect. If we were to find a frame in which
> the vacuum speed of light was not c, would this not invalidate the
> postulates of Relativity, and therefore disprove it?

A physical theory makes a claim about the way the world works. It
can be falsified by experiment. If you want to use the word "disprove"
instead of "falsify", that's fine, as long as it's clear what you
mean.

A *theorem* is not a claim about the way the world works, it is
a claim about the *consequences* of certain assumptions. Bell is
not making a claim about the world, he's making a claim about a
certain class of theories--if those theories are true, then certain
things follow from that.

You can attack such a theorem in several ways: (1) You can show that
it's logically invalid--it doesn't actually prove what it purports
to prove. (2) You can show (and you could consider this a special
case of (1)) that there are implicit assumptions relied on in the
proof that were not made explicit. Usually, discovering such hidden
assumptions is a first step toward the third way: (3) You can show
that the theorem is not applicable, because the assumptions don't
hold (or may not hold) in cases of interest.

> For example, Einstein made an initial estimate for the angular
> distortion of light for Eddington to look for. He later corrected
> this. Did Einstein disprove Relativity with the correction of a
> mathematical error? What you say above seems to say so.

There are two different activities that went on: (1) the formulation of
a physical theory, GR, that made testable claims about the world, (2) a
purely mathematical calculation of the consequences of that theory.

Einstein made a mistake in the mathematics. That doesn't disprove
the physical theory.

[harald]

"Daryl McCullough" <stevend...@yahoo.com> wrote in message

news:3458321.3037.1328532291489.JavaMail.geo-discussion-forums@vbuf18...

> On Monday, February 6, 2012 6:59:36 AM UTC-5, harald wrote:
>> "Daryl McCullough" <stevend...@yahoo.com> wrote in message
>> news:27069963.147.1324041602008.JavaMail.geo-discussion-forums@vbbhx10...
>> > On Thursday, December 15, 2011 1:18:44 PM UTC-5, mpc755 wrote:
>> > You don't disprove a theorem except by
>> > discovering a mathematical error in it.
>>
>> That is true for a mathematical theorem. However, Bell's is a theorem of
>> physics: it's a claim about physical theories, based on physical
>> assumptions. Already the term "local realistic theory" does not belong to
>> mathematics.
>> Compare: http://plato.stanford.edu/entries/bell-theorem/
>
> As I said, I don't think that the word "disproof" can be meaningfully
> applied to theorems.

Probably you simply disagree with the use of the word "theorem" for Bell's
claim about QM which happens to be called "Bell's Theorem".

> You can attack Bell's theorem on several grounds:
> you can find a mistake in the derivation, or you can find a hidden
> assumption that should be made explicit, or you can argue that assumptions
> (explicit or implicit) don't apply in cases of interest, or you can argue
> that Bell's formalization of the problem (the way he tries to capture
> physical concepts such as "locality" and "realism") is not correct.
>
> The terminology, "disproving Bell's Theorem" is not very
> descriptive of any of these lines of attack.

I understand "disproving Bell's Theorem" to simply mean that Bell's claim
("theorem") was based on false reasoning or false assertions; which claim
was that no physical theory which is realistic and also local in a specified
sense can agree with all of the statistical implications of Quantum
Mechanics. (Bell's Theorem according to Abner Shimony.)

[Daryl McCullough]

On Monday, February 6, 2012 10:34:00 AM UTC-5, harald wrote:
> "Daryl McCullough" <stevend...@yahoo.com> wrote

> > You can attack Bell's theorem on several grounds:
> > you can find a mistake in the derivation, or you can find a hidden
> > assumption that should be made explicit, or you can argue that assumptions
> > (explicit or implicit) don't apply in cases of interest, or you can argue
> > that Bell's formalization of the problem (the way he tries to capture
> > physical concepts such as "locality" and "realism") is not correct.
> >
> > The terminology, "disproving Bell's Theorem" is not very
> > descriptive of any of these lines of attack.
>
> I understand "disproving Bell's Theorem" to simply mean that Bell's claim
> ("theorem") was based on false reasoning or false assertions;

I assume that by "false reasoning" you mean the same thing as
a mistake in the proof. In that case, a better description is
to say: "Mistake discovered in Bell's proof" rather than "Bell's
theorem disproved".

As for "false assertions", I think that it's important to
distinguish the various ways that the proof can be inapplicable.

> which claim was that no physical theory which is realistic and
> also local in a specified sense can agree with all of the
> statistical implications of Quantum Mechanics. (Bell's Theorem
> according to Abner Shimony.)

That's not a false assertion, unless there is a mistake in Bell's
proof.

[Joy Christian]

On Feb 6, 2:04 pm, Daryl McCullough <stevendaryl3...@yahoo.com> wrote:

>
> The terminology, "disproving Bell's Theorem" is not very
> descriptive of any of these lines of attack.
>

Bell claimed that statistically independent local functions
of the form

A(a, L) = +1 or -1 and B(b, L) = +1 or -1

cannot reproduce correlation of the form

< AB > = -a.b.

"This is the theorem", to quote Bell's exact words.

That this "theorem" of Bell is mathematically false is clear
from this one-page paper: http://arxiv.org/abs/1103.1879.
In this paper I have constructed two purely local functions,
A and B, that reproduce the correlation < AB > = -a.b.

Thus I have decisively *disproved* Bell's mathematical
theorem -- whatever the physical claim he intended.

Joy Christian

[Daryl McCullough]

On Monday, February 6, 2012 1:27:54 PM UTC-5, Joy Christian wrote:

> Bell claimed that statistically independent local functions
> of the form
>
> A(a, L) = +1 or -1 and B(b, L) = +1 or -1
>
> cannot reproduce correlation of the form
>
> < AB > = -a.b.
>
> "This is the theorem", to quote Bell's exact words.
>
> That this "theorem" of Bell is mathematically
> false is clear from this one-page paper:
> http://arxiv.org/abs/1103.1879.

You keep using the word "clear". I don't think you
use in the usual sense of the word.

You did not disprove his theorem. You proved a
related theorem that does not contradict his
theorem.

Bell proved that the nonexistence of
A, B, rho and V with the following
properties:

(1) V is a set.
(2) rho is a real-valued probability distribution
on V.
(3) A and B are functions that takes a pair (a,v)
where a is a unit 3-vector, and v is an element
of V, and returns an element of the set {-1,+1}.
(4) For fixed a and b, <A B> = - cos(theta)
where <A B> is defined to be:
Sum over all v in V of A(a,v) B(b,v) rho(v),
and where theta is the angle between a and b.

You did not disprove this theorem, and it is
extremely misleading for you to claim otherwise.

[Daryl McCullough]

On Monday, February 6, 2012 3:05:43 PM UTC-5, Daryl McCullough wrote:

> Bell proved that the nonexistence of
> A, B, rho and V with the following
> properties:
>
> (1) V is a set.
> (2) rho is a real-valued probability distribution
> on V.
> (3) A and B are functions that takes a pair (a,v)
> where a is a unit 3-vector, and v is an element
> of V, and returns an element of the set {-1,+1}.
> (4) For fixed a and b, <A B> = - cos(theta)
> where <A B> is defined to be:
> Sum over all v in V of A(a,v) B(b,v) rho(v),
> and where theta is the angle between a and b.
>
> You did not disprove this theorem, and it is
> extremely misleading for you to claim otherwise.

Sorry, what I said was not correct. Bell proved
an inequality that was satisfied by any A, B,
rho, and V, and he showed that the result

<A B> = - cos(theta)

violated that inequality. So he showed something
more general than the nonexistence of something
satisfying

[Joy Christian]

On Feb 6, 8:05 pm, Daryl McCullough <stevendaryl3...@yahoo.com> wrote:

> You keep using the word "clear". I don't think you
> use in the usual sense of the word.

English may not be my first language but I think
I do understand the meaning of the word "clear."

It is clear from this paper http://arxiv.org/abs/1103.1879
that the form of functions considered by Bell in the very
first equation of his famous paper is no different from
the form of functions A(a, L) and B(b, L) defined in the
equations (1) and (2) of the linked paper. Moreover, it
is clear from equation (7) of the linked paper that the
correlation between A and B is precisely < AB > = -a.b.

Now Bell explicitly states in several chapters of his book
that this is impossible. In particular, he explicitly states,
and I quote, "This is the theorem." You may dress this
up differently from Bell's own words, but you cannot deny
the fact that I have explicitly disproved Bell's mathematical
claim, as stated by him in several chapters of his book.

But let us play the game your way. Let us go through
your checklist:

> (1) V is a set.

Check!

> (2) rho is a real-valued probability distribution on V.

Check!

> (3) A and B are functions that takes a pair (a,v) where a
> is a unit 3-vector, and v is an element of V, and returns
> an element of the set {-1,+1}.

Check!

> (4) For fixed a and b, <A B> = - cos(theta)
> where <A B> is defined to be:

> sum over all v in V of A(a,v) B(b,v) rho(v),

> and where theta is the angle between a and b.

Check!

I see absolutely no difference between this and what
I have considered. You can check for yourself if you
wish. Go through equations (1.33) and (1.50) of this
paper http://arxiv.org/abs/1201.0775, for example, and
tell me how, or where, does my disproof differ from your
version of Bell's theorem. And while you are at it, do let us
know why the difference -- if you find any -- is physically
relevant for the EPR argument and the observed results.

It quite disingenuous to keep looking for flimsy excuses
and cooked up flaws in my work to deny the obvious fact
that I have decisively disproved Bell's theorem.

Joy Christian

[Joy Christian]

On Feb 6, 9:26 pm, Daryl McCullough <stevendaryl3...@yahoo.com> wrote:

> Sorry, what I said was not correct. Bell proved
> an inequality that was satisfied by any A, B,
> rho, and V, and he showed that the result
> <A B> = - cos(theta)
> violated that inequality. So he showed something
> more general than the nonexistence of something
> satisfying
> <A B> = - cos(theta)

But I have proved something far more general than
even this. See for example the derivations in the
following two papers:

http://arxiv.org/abs/0904.4259

and

http://arxiv.org/abs/1101.1958

And even more general treatment for ALL quantum
correlations can be found in this paper:

http://arxiv.org/abs/1201.0775

I have gone far beyond Bell;s theorem and uncovered
the true raison d'être of ALL quantum correlations.

Joy Christian

[Daryl McCullough]

On Monday, February 6, 2012 6:43:25 PM UTC-5, Joy Christian wrote:

> It is clear from this paper http://arxiv.org/abs/1103.1879
> that the form of functions considered by Bell in the very
> first equation of his famous paper is no different from
> the form of functions A(a, L) and B(b, L) defined in the
> equations (1) and (2) of the linked paper. Moreover, it
> is clear from equation (7) of the linked paper that the
> correlation between A and B is precisely < AB > = -a.b.

But you are using a different integral to calculate <AB>.
Bell is claiming that it is impossible to have

(Integral of rho(v) A(a,v) B(b,v) dv) = -a.b

And that's true.

> Now Bell explicitly states in several chapters of his book
> that this is impossible. In particular, he explicitly states,
> and I quote, "This is the theorem." You may dress this
> up differently from Bell's own words, but you cannot deny
> the fact that I have explicitly disproved Bell's mathematical
> claim, as stated by him in several chapters of his book.

No, you didn't. You proved a related theorem, but it was not
one that contradicted Bell's theorem.

I don't see how your statements about "different scales of dispersion"
makes any sense. In the usual statistical analysis, the reason for
dividing by the standard deviation is for the purpose of making a
measure of correlation that is independent of scale and zeros of
the two functions A and B. It's a standard way to compare the
strengths of a linear relationship. There is no law that one must
divide by the standard deviation.

But in the case of the functions script-A and script-B, the standard
deviation is *already* equal to 1! The definition of standard
deviation is this: standard-deviation(A) = square-root(variance(A)),
where variance(A) = <A^2> - <A>^2. In our case, A^2 = 1, and
<A> = 0, so variance of A is *already* equal to 1.

> > (4) For fixed a and b, <A B> = - cos(theta)
> > where <A B> is defined to be:
> > sum over all v in V of A(a,v) B(b,v) rho(v),
> > and where theta is the angle between a and b.

Check!

> I see absolutely no difference between this and what
> I have considered.

The difference is that in your integral, you are dividing
by a Clifford-valued operator that you are calling the
"standard deviation". I don't think that it makes any sense
for the standard deviation to be a Clifford number. That is
not a "standard" notion of standard deviation.

I don't see that what you have done makes any sense. It certainly
isn't a disproof of Bell's theorem.

[Joy Christian]

On Feb 7, 5:39 am, Daryl McCullough <stevendaryl3...@yahoo.com> wrote:

> But you are using a different integral to calculate <AB>.

What different integral? I am using the integral, or sum,
defined by Newton and Leibniz which I learned in my
calculus class in high-school. Is Bell and you are using
something else – some different calculus than Newton's?

> Bell is claiming that it is impossible to have
>
> (Integral of rho(v) A(a,v) B(b,v) dv) = -a.b
>
> And that's true.

I have decisively shown that this claim of Bell is false.
Please go through my claulations in any of my papers,
for example this one, http://arxiv.org/abs/1201.0775, to
recognize that I have disproved Bell's claim.

> > Now Bell explicitly states in several chapters of his book
> > that this is impossible. In particular, he explicitly states,
> > and I quote, "This is the theorem." You may dress this
> > up differently from Bell's own words, but you cannot deny
> > the fact that I have explicitly disproved Bell's mathematical
> > claim, as stated by him in several chapters of his book.
>
> No, you didn't.

Yes, I did. Flat denial of facts is not an argument.

> You proved a related theorem, but it was not
> one that contradicted Bell's theorem.

Nonsense. I have decisively *disproved* Bell's theorem (as
he has repeatedly stated in his book and elsewhere) for
example in this paper: http://arxiv.org/abs/1201.0775

> I don't see how your statements about "different scales
> of dispersion" makes any sense.

Are you claiming that the statistical procedure discovered
by Galton and Pearson over a century ago and used around
the world to calculate correlations makes no sense? Have you
and Bell invented a special statistical method for yourselves?

> In the usual statistical analysis, the reason for
> dividing by the standard deviation is for the purpose of making a
> measure of correlation that is independent of scale and zeros of
> the two functions A and B. It's a standard way to compare the
> strengths of a linear relationship.

You said it. Thank you. I am using *the* standard way to compare
the strength of linear relationship. Thank you.

> There is no law that one must divide by the standard deviation.

There is no law against comparing apples and oranges either.
Without standardizing or normalizing the raw scores one cannot
copare them. It would be like exchanging dollars for yens without
using the conversion factor. I have some Japanese yens. I sure
would like to trade with your US dollars, given your sense of basic
statistics.

> But in the case of the functions script-A and script-B, the standard
> deviation is *already* equal to 1!

Nonsense. You cannot replace my model with a straw man of your
own and then try to knock it off. I understand that that is the only
strategy left for you, but it also happens to be a logical fallacy.
I urge the readers to please read my papers to see what my
model is actually saying: http://arxiv.org/abs/1106.0748

> > > (4) For fixed a and b, <A B> = - cos(theta)
> > >  where <A B> is defined to be:
> > > sum over all v in V of A(a,v) B(b,v) rho(v),
> > > and where theta is the angle between a and b.
>
> Check!
>
> > I see absolutely no difference between this and what
> > I have considered.
>
> The difference is that in your integral, you are dividing

> by a Clifford-valued operator...

Nonsense. There are no "operators", either in my work,
or in geometric algebra in general. Geometric algebra is
a completion of a *real* number system. If you are not
familiar with the language of geometric algebra then the
honest thing to do is to admit that fact rather than try to
misread and misrepresent my work to suit your ideology.

> I don't see that what you have done makes any sense.

This is just a flimsy excuse, based on a straw man fallacy,
for denying a clear, straightforward, physically insightful,
and decisive refutation of Bell's theorem.

Joy Christian

[harald]

"Daryl McCullough" <stevend...@yahoo.com> wrote in message

news:3700533.609.1328548238803.JavaMail.geo-discussion-forums@vbiq27...

> On Monday, February 6, 2012 10:34:00 AM UTC-5, harald wrote:
>> "Daryl McCullough" <stevend...@yahoo.com> wrote

[..]
>> [Bell's] claim was that no physical theory which is realistic and

>> also local in a specified sense can agree with all of the
>> statistical implications of Quantum Mechanics. (Bell's Theorem
>> according to Abner Shimony.)
>
> That's not a false assertion, unless there is a mistake in Bell's
> proof.

Yes, that's what Christian claims - and we are still not convinced one way
or the other.

[Daryl McCullough]

On Tuesday, February 7, 2012 2:37:48 AM UTC-5, Joy Christian wrote:

> Are you claiming that the statistical procedure discovered
> by Galton and Pearson over a century ago and used around
> the world to calculate correlations makes no sense?

No, I'm claiming that you are misusing it in a way that makes
no sense. There are three misunderstandings involved. First, the
notion of "standard deviation" is appropriate only for *real-valued*
random variables. It doesn't make a bit of sense to talk about
a standard deviation being a Clifford number. Secondly, the
standard deviation for any random variable that takes on values
+/- 1, and has mean 0 is 1. Not a Clifford number.

Third, the point of the standard formula for correlation is to
measure the strength of a linear relationship. It's a *heuristic*
measurement to discover patterns in statistical data.

> > In the usual statistical analysis, the reason for
> > dividing by the standard deviation is for the purpose of making a
> > measure of correlation that is independent of scale and zeros of
> > the two functions A and B. It's a standard way to compare the
> > strengths of a linear relationship.
>
> You said it. Thank you. I am using *the* standard way to compare
> the strength of linear relationship. Thank you.

No, you're not using in the standard way, at all. For one thing,
the relationship between A and B is not linear. For another thing,
you're not attempting to *discover* a relationship. We know precisely
what the relationship is between A and B. What we're doing instead
is trying to develop a theory that accurately describes that relationship.

In particular, what the point of a hidden variable theory of the type
Bell was investigating is to account for both the nondeterminism in
the measured results in the two detectors, and for their correlation.
Your formulas *don't* do that. They *don't* account for the nondeterminism
in the results.

> > There is no law that one must divide by the standard deviation.
>
> There is no law against comparing apples and oranges either.
> Without standardizing or normalizing the raw scores one cannot

> compare them.

That's just nonsense. Alice is performing an experiment, and
based on the outcome, she writes down +1 or -1 as the result.
Bob is similarly performing an experiment, and similarly writes
down +1 or -1. I can certainly ask the question: Out of 10
runs of the experiment, how many times did they both write down
+1, how many times did they both write down -1, how many times
did they write down the opposite number?

It's a perfectly reasonable thing to ask. And experimentally,
the answer is that if Alice holds her orientation fixed at
direction a, and Bob holds his orientation fixed at direction
b, then for a large number of trials (let's assume that the
angle between a and b is held fixed at 120 degrees, and that
we're doing the spin-1/2 twin pair experiment):

1/8 of the time, Alice and Bob both get +1.
1/8 of the time, Alice and Bob both get -1.
3/8 of the time, Alice gets +1, Bob gets -1.
3/8 of the time, Alice gets -1, Bob gets +1.

That's the experimental data that is to be explained
by a hidden-variable explanation. To successfully
explain this data using a *deterministic* hidden-variable
model would require at the least two functions A(a,mu),
B(b,mu), and a sequence of values mu_1, mu_2, etc.
such that the nth outcome for Alice is equal to
A(a,mu_n) and the nth outcome for Bob is equal to
B(b,mu_n). Your proposed functions A(a,mu) B(b,mu)
can't possibly do that.

> It would be like exchanging dollars for yens without
> using the conversion factor.

If experimentally, I find a number of yen floating in
the Pacific Ocean in Japan, and on the same day, I discover a
number of dollars in the Atlantic Ocean, then I can certainly
compare the number of yen in the Pacific with the number of
dollars in the Atlantic without knowing how many yen are worth
a dollar.

What you're saying doesn't make sense to me.

[Joy Christian]

On Feb 7, 3:45 pm, Daryl McCullough <stevendaryl3...@yahoo.com> wrote:

> No, I'm claiming that you are misusing it in a way that makes no sense.

I am not misusing anything. I am using bread and butter
geometric algebra. Are you claiming that geometric algebra
makes no sense? If so, then your dispute is not with me.

> There are three misunderstandings involved.

Misunderstandings, if there are any, are entirely yours.

> the notion of "standard deviation" is appropriate only for *real-valued*
> random variables. It doesn't make a bit of sense to talk about
> a standard deviation being a Clifford number.

A bivector (or even a multi-vector) is a *real-valued* variable.
If you do not understand even such a basic fact of geometric
algebra then you indeed have a long way to go.

Besides, statistical concepts such as expectation values and
standard deviations are routinely employed for complex valued
and hyper-complex valued random numbers. Your lack of knowledge
about such basic facts is your problem, not mine.

> standard deviation for any random variable that takes on values
> +/- 1, and has mean 0 is 1.

Not so in my model. Have a look at my papers. Perhaps you
will learn a thing or two about geometric algebra.

> > > In the usual statistical analysis, the reason for
> > > dividing by the standard deviation is for the purpose of making a
> > > measure of correlation that is independent of scale and zeros of
> > > the two functions A and B. It's a standard way to compare the
> > > strengths of a linear relationship.
>
> > You said it. Thank you. I am using *the* standard way to compare
> > the strength of linear relationship. Thank you.
>
> No, you're not using in the standard way, at all.

I completely disagree. I AM using it in a perfectly standard way.
If you do not see this, then it can only be because you have no
understanding of geometric algebra. In fact it is evident from your
assertions that you do not know the first thing about it.

> For one thing, the relationship between A and B is not linear.

Indeed. And therefore Bell's analysis is a non-starter. Because he
starts with the assumption that the relationship between A and B
is linear, and then proves that it must be linear. In other words,
Bell
assumes from the start what he sets out to prove. It is all a big
circle.

> you're not attempting to *discover* a relationship. We know precisely
> what the relationship is between A and B. What we're doing instead
> is trying to develop a theory that accurately describes that relationship.

The reason the relationship between A and B is sinusoidal is because
A and B represent two equatorial points of a parallelized 3-sphere,
which
in turn correctly encodes the observed symmetries of our physical
space.

> In particular, what the point of a hidden variable theory of the type
> Bell was investigating is to account for both the nondeterminism in
> the measured results in the two detectors, and for their correlation.

This is gobbledygook. You are injecting theoretical preconceptions
such as "non-determinism" into empirical facts such as measurement
results. Measurement results are measurement results. They do not
have any non-determinism in them other than the usual classical
randomness.You have zero understanding of the EPR-Bell debate.
It had nothing to do with non-determinism. Bell went out of his way
to stress that, and took pains to clear up precisely the kind of
misconceptions you are exhibiting.

> > > There is no law that one must divide by the standard deviation.
>
> > There is no law against comparing apples and oranges either.
> > Without standardizing or normalizing the raw scores one cannot
> > compare them.
>
> That's just nonsense.

You have just called the deep statistical insights, painfully arrived
at by people like Galton and Pearson, "nonsense." Your dispute is
thus not with me but with them. I accept their valuable contributions.

> Alice is performing an experiment, and
> based on the outcome, she writes down +1 or -1 as the result.
> Bob is similarly performing an experiment, and similarly writes
> down +1 or -1. I can certainly ask the question: Out of 10
> runs of the experiment, how many times did they both write down
> +1, how many times did they both write down -1, how many times
> did they write down the opposite number?

If only things were that simple. The real experiments are never that
simple. And besides, one only has to look at the actual practice of
experimenters to note that without normalization or standardization
comparison between raw scores is meaningless. Moreover, the numbers
A and B in my **theoretical** model are generated with built-in
standard
deviations. It is then simply incorrect to compare the raw scores
generated by them without standardizing the variables first. This is
basic, high-school statistics.

> It's a perfectly reasonable thing to ask. And experimentally,
> the answer is that if Alice holds her orientation fixed at
> direction a, and Bob holds his orientation fixed at direction
> b, then for a large number of trials (let's assume that the
> angle between a and b is held fixed at 120 degrees, and that
> we're doing the spin-1/2 twin pair experiment):
>
> 1/8 of the time, Alice and Bob both get +1.
> 1/8 of the time, Alice and Bob both get -1.
> 3/8 of the time, Alice gets +1, Bob gets -1.
> 3/8 of the time, Alice gets -1, Bob gets +1.
>
> That's the experimental data that is to be explained
> by a hidden-variable explanation.

You will find the explanation you seek in my papers:
http://arxiv.org/find/all/1/au:+Christian_Joy/0/1/0/all/0/1

> To successfully
> explain this data using a *deterministic* hidden-variable
> model would require at the least two functions A(a,mu),
> B(b,mu), and a sequence of values mu_1, mu_2, etc.
> such that the nth outcome for Alice is equal to
> A(a,mu_n) and the nth outcome for Bob is equal to
> B(b,mu_n). Your proposed functions A(a,mu) B(b,mu)
> can't possibly do that.

This is quite funny. The evidence is right in front of you. It is
all there in my papers. And yet you are flatly denying the
evidence. This reminds me of those poor souls who, despite
witnessing multiple demonstrations by the Wright brothers,
continued to insist that "heavier-than-air flying machines
are impossible -- what the lying brothers are claiming must
therefore be a conjuring trick. It simply does not make sense
to fly a heavier-than-air machine."

> What you're saying doesn't make sense to me.

It never will.

While you stay put firmly on ground, I am off flying
my flying machine: http://arxiv.org/abs/1201.0775

Joy Christian

[Daryl McCullough]

On Tuesday, February 7, 2012 4:15:56 PM UTC-5, Joy Christian wrote:

> You have just called the deep statistical insights, painfully arrived
> at by people like Galton and Pearson, "nonsense." Your dispute is
> thus not with me but with them. I accept their valuable contributions.

I don't have a dispute with the use of standard deviations in statistics.
I don't have a dispute with the use of geometric algebra.
But I don't see any meaningful way in which a standard deviation can
be a bivector. Can you point to a reference where a researcher
considers bivector standard deviations?

[Joy Christian]

Yes.

Here are two references by a researcher
(with pointers to other references on GA)
who considers bivector standard deviations:
http://arxiv.org/abs/1106.0748 and
http://arxiv.org/abs/1201.0775

Joy Christian

[harald]

"Joy Christian" <hojo...@gmail.com> wrote in message
news:59b68b91-4660-4c9a...@t15g2000yqi.googlegroups.com...

What was obviously meant: can you point to a reference where an other
researcher than yourself considers bivector standard deviations? Indeed,
that would be helpful!

Harald

[Joy Christian]

On Feb 8, 10:14 am, "harald" <h...@swissonline.ch> wrote:


> What was obviously meant: can you point to a reference where another
> researcher than yourself considers bivector standard deviations? Indeed,
> that would be helpful!

Your request is equivalent to requesting Kepler to point
out another astronomer who considers elliptical orbits.


Joy Christian

[Heine Rasmussen]

Temporarily putting the question whether you are the new Kepler or not
aside, it would still be very helpful to us if you could provide us
with a theoretical explanation of the theory of bivector standard
deviations, or at least give a pointer to an independent exposition of
such. In your papers, this is not sufficiently elaborated upon, and
the reader is left with a feeling that "here, a miracle happens, so we
conclude that..."

[Joy Christian]

On Feb 8, 8:41 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> Temporarily putting the question whether you are the new Kepler or
> not aside, it would still be very helpful to us if you could provide us
> with a theoretical explanation of the theory of bivector standard
> deviations, or at least give a pointer to an independent exposition of
> such.  In your papers, this is not sufficiently elaborated upon, and
> the reader is left with a feeling that "here, a miracle happens, so we
> conclude that..."

To my mind bivectorial standard deviation is not all
that difficult to understand once you appreciate the
fact that a bivector is a real-valued number, just like
any other real-valued number we are familiar with.
You can verify this statement in any textbook on
geometric algebra, references to which can be found
in almost all of my recent papers.

To see this, let us consider two bivectors, w and w(k)
where k = +1 or -1 is a random variable and w(k) = k w,
with w^2 = -1. It is important to recognize that the latter
property DOES NOT make the bivector “complex.” It is
a purely geometrical entity, with an unfamiliar property.

Next, let us define a variable A = w w(k). Since w is a
constant number, errors generated within A by the random
process k stem entirely from the random variable w(k) and
propagate linearly. In other words, the standard deviation of
the random number A due to the process k is given by

sigma(A) = w x sigma[w(k)]

It is now easy to check using the properties of w specified
above (namely the fact that w is a unit bivector) that

sigma[w(k)] = 1.

>From the above equations it is then clear that

sigma(A) = w

Thus the typical error generated within A, due to the
random process k, is w. Therefore the standardized
variable which must be used to compare the raw score
A with some other raw score (such as B) is A / w = w(k).

That is all there is to it.

Well, almost. It helps to know that in geometric algebra
scalars such as A and bivectors such as w are all treated
on equal footing. They are all numbers, but of different
grade. It is a graded algebra of numbers like A and w.

Joy Christian

[Ken S. Tucker]

I tend to agree with you (Heine) as Harald introduced,
I usually toss "bivectors" into the asymmetrical tensor bin,
however some other uses of the term are here,

http://en.wikipedia.org/wiki/Bivector

in vectors ,dyads, triads, etc.
Perhaps Joy might point to the best definition we should study.
Regards
Ken S. Tucker

http://en.wikipedia.org/wiki/Bivector

[FrediFizzx]

"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:4bb9078f-957e-4b8b...@jn12g2000pbb.googlegroups.com...

As used in Geometric Algebra. The Wikipedia article looks OK. Ken, too bad
you are on slow dial up because the other part of Joy's model is
understanding the 3-sphere topology and there is a good video lecture about
that here,

http://www.youtube.com/watch?v=GQB4DjEI_LI
Niles Johnson; "Visualizations of the Hopf fibration

Best,

Fred Diether

[Ken S. Tucker]

On Feb 8, 11:42 pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
> "Ken S. Tucker" <dynam...@vianet.on.ca> wrote in messagenews:4bb9078f-957e-4b8b...@jn12g2000pbb.googlegroups.com...

Apparently science is communicated more via "youtube" recently.
I suspect the need for bivectors herein is unnecessary.
Ken

[Heine Rasmussen]

On 9 Feb, 00:35, Joy Christian <hojoin...@gmail.com> wrote:
[...]

>
> To see this, let us consider two bivectors, w and w(k)
> where k = +1 or -1 is a random variable and w(k) = k w,
> with w^2 = -1. It is important to recognize that the latter
> property DOES NOT make the bivector “complex.” It is
> a purely geometrical entity, with an unfamiliar property.
>
> Next, let us define a variable A = w w(k). Since w is a
> constant number, errors generated within A by the random
> process k stem entirely from the random variable w(k) and
> propagate linearly. In other words, the standard deviation of
> the random number A due to the process k is given by
>
>             sigma(A)  =  w x sigma[w(k)]
>
> It is now easy to check using the properties of w specified
> above (namely the fact that w is a unit bivector) that
>
>                       sigma[w(k)] = 1.
>
> >From the above equations it is then clear that
>
>                        sigma(A) = w

[...]

But I was rather thinking about your particular definition of sigma.
In arXiv:1106.0748v3, eqn. (24), you define sigma to be the square
root of the average of squared norms, which I have no problems with,
as this is the canonical way to generalize standard deviation to
anything that has a norm. Note that this will always make sigma a non-
negative real scalar, or grade-0 in GA language.

The sentence that doesn't make sense to me, is then “Since w is a

constant number, errors generated within A by the random process k

stem entirely from the random variable w(k) and propagate linearly”.
So you appearantly move w outside the norm operator, and since the
remaining expression evaluates to unity, the standard deviation now
becomes a bivector (grade-2). I don't see how you can do that, since
this seems not to be consistent with the definition of sigma in eqn
(24).

[Joy Christian]

On Feb 11, 2:08 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> But I was rather thinking about your particular definition of sigma.
> In arXiv:1106.0748v3, eqn. (24), you define sigma to be the square
> root of the average of squared norms, which I have no problems with,
> as this is the canonical way to generalize standard deviation to
> anything that has a norm.  Note that this will always make sigma a

> non-negative real scalar, or grade-0 in GA language.

>
> The sentence that doesn't make sense to me, is then “Since w is a
> constant number, errors generated within A by the random process k
> stem entirely from the random variable w(k) and propagate linearly”.
> So you appearantly move w outside the norm operator, and since the
> remaining expression evaluates to unity, the standard deviation now
> becomes a bivector (grade-2).  I don't see how you can do that, since
> this seems not to be consistent with the definition of sigma in eqn (24).

There are several ways to see the consistency of what I am
doing. The problem is to ascertain the typical error within A
due to the random process k = +1 or -1, where A is a polar
point of the parallelized 3-sphere (either a north pole or a
south pole) and k is a fair coin. Now before checking the
consistency of the formulae for sigma it is instructive to first
check out the geometrical picture of what is going on. Since
A is a geometric product of a fixed bivector w and a random
bivector w(k) = k w, and since w^2 = -1, we see that A = w w(k)
= +1 (i.e., "north" pole) when w(k) = -w, and A = w w(k) = -1 (i.e.,
"south" pole) when w(k) = +w. Thus a typical error within A is
clearly w, not 1. Moreover, since we are working within the even
subalgebra of the Clifford algebra Cl(3, 0) (which is represented
by the sum of a scalar and a bivector), it is not surprising to see
that the typical error within a scalar is a bivector (namely w) and
the typical error within a bivector is a scalar (namely 1).

Now what about the formulae for sigma? Well, they are
perfectly consistent too. It seems to me that you have
missed the difference between A and script-A appearing
in my equation (23). The key equation to understand what
I am doing is equation (23), not (24). The latter, as you note,
is the canonical way to calculate the standard deviation
within GA. It is evident from equation (23) that I am not
bringing anything out of the norm operator in equation (24).
The bivector (-I . a) has never been inside the norm operator
to being with. It is outside of sigma(A) from the outset, as noted
in equation (23), which is equivalent to the Leibniz rule within
the even subalgebra of Cl(3,0). Deviations in script-A are thus
simply (-I . a) (or w) times the deviations in A.

Joy Christian

[Heine Rasmussen]

>[...]
> Joy Christian

Then I suggest that you provide us with a formal definition of
standard deviation (beyond the phrase "typical error") that shows how
the standard deviation of a random scalar can come out as a bivector.

[Joy Christian]

On Feb 15, 2:07 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> Then I suggest that you provide us with a formal definition of
> standard deviation (beyond the phrase "typical error") that shows how
> the standard deviation of a random scalar can come out as a bivector.

Let A = f(w) = v w be a random scalar generated by a geometric
product of two bivectors, v and w, where v is a constant bivector
and w is a random bivector of constant norm. Suppose further that
w is normally distributed with mean m(w) and standard deviation
sigma(w) [you have already agreed with my definition of sigma(w)].

Now we would like to know how the 68% probability interval

[ m(w) - sigma(w), m(w) + sigma(w) ]

is propagated from the random bivector w to the random scalar A
through the function f(w) = v w. Now from the definition of A and a
first-order Taylor expansion of f(w) about the point w = m(w) we have

A = f [m(w)] + [ d f / d w ] [ w - m(w) ] + ...

But it is easy to see that [ d f / d w ] = v = constant. Therefore
the parameters of propagation of the 68% probability interval are

m(A) = f [m(w)]

and

sigma(A) = [ d f / d w ] x sigma(w)

= v x sigma(w).

The desired 68% probability interval of the distribution of the
scalar A is thus given by

[ m(A) - sigma(A), m(A) + sigma(A) ].

If now we assume that w is a unit bivector with vanishing mean
and unit norm (as in my papers), then we have

m(A) = 0 and sigma(A) = v.

Any remaining problems are purely psychological. We are
not used to seeing bivectors treated as numbers. But they
are nothing but numbers, albeit with some unusal properties.
What is more, as I have already noted, in geometric algebra
scalars, vectors, bivectors, and trivectors are all treated on
equal footing. It is a truly democratic algebra.

Joy Christian

[Ken S. Tucker]

To Heine, Joy et al...

IMVHO (In My Very Humble Opinion), Joys use of bivectors
applied to probability is a ' flash of genius'.
((My background also includes insurance and WW3 sims)).

Joy, perhaps you might understand the unique application
of bivectors to probablities is new to some of us.
I believe that's a 'measurement issue'.
Regards
Ken S. Tucker

[FrediFizzx]

"Joy Christian" <hojo...@gmail.com> wrote in message

news:bfaff71d-cffc-4b9d...@gr6g2000vbb.googlegroups.com...

And.... IMHO leads to new cutting edge physical insights.

"On the Origins of Quantum Correlations"
http://arxiv.org/abs/1201.0775

Best,

Fred Diether

[Joy Christian]

On Feb 15, 7:57 pm, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:

> IMVHO (In My Very Humble Opinion), Joys use of bivectors
> applied to probability is a ' flash of genius'.
> ((My background also includes insurance and WW3 sims)).
>
> Joy, perhaps you might understand the unique application
> of bivectors to probablities is new to some of us.
> I believe that's a 'measurement issue'.
> Regards
> Ken S. Tucker

Thanks, Ken; thanks, Fred.

I am constantly learning that what is obvious to me
may not necessarily be obvious to everyone else.

And you are right, Ken, it is a "measurement issue."

Or more precisely, the so-called measurement problem
(or even Born rule) has no relevance within my framework.

Joy Christian

[Heine Rasmussen]

On 15 Feb, 19:15, Joy Christian <hojoin...@gmail.com> wrote:
> On Feb 15, 2:07 pm, Heine Rasmussen <hein...@gmail.com> wrote:
>
> > Then I suggest that you provide us with a formal definition of
> > standard deviation (beyond the phrase "typical error") that shows how
> > the standard deviation of a random scalar can come out as a bivector.
>
> Let A = f(w) = v w be a random scalar generated by a geometric
> product of two bivectors, v and w, where v is a constant bivector
> and w is a random bivector of constant norm. Suppose further that
> w is normally distributed with mean m(w) and standard deviation
> sigma(w) [you have already agreed with my definition of sigma(w)].

And how could a random bivector of constant norm be normally
distributed?

> Now we would like to know how the 68% probability interval
>
>                  [ m(w) - sigma(w),  m(w) + sigma(w) ]

Intervals have no meaning in GA exept for the scalar subspace. A
Clifford algebra is not a well ordered set.

> is propagated from the random bivector w to the random scalar A
> through the function f(w) = v w. Now from the definition of A and a
> first-order Taylor expansion of f(w) about the point w = m(w) we have
>
>               A  =  f [m(w)]  +  [ d f / d w ] [ w - m(w) ]  +  ...
>
> But it is easy to see that [ d f / d w ]  =  v = constant. Therefore
> the parameters of propagation of the 68% probability interval are
>
>                              m(A)  =  f [m(w)]

At least this is true.

> and
>
>                   sigma(A)  =  [ d f / d w ] x sigma(w)
>
>                                  =  v x sigma(w).

Since your interval is not well defined, this is not a valid
inference.

[Joy Christian]

On Feb 16, 4:59 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> And how could a random bivector of constant norm be normally
> distributed?

Here is how: A bivector has three properties: (1) magnitude,
(2) direction, and (3) a sense of rotation. Random bivectors
of constant norm thus have two unspecified properties: their
direction and their sense of rotation. The set of all such random
bivectors represent equatorial points of a parallelized 3-sphere,
whose orientation (or local handedness) remains unspecified.
A Gaussian distribution of such bivectors is thus a normal
probability distribution of these unspecified properties. It is a
perfectly well defined scalar function of w, m(w), and sigma(w).

> > Now we would like to know how the 68% probability interval
>
> >                  [ m(w) - sigma(w),  m(w) + sigma(w) ]
>
> Intervals have no meaning in GA exept for the scalar subspace. A
> Clifford algebra is not a well ordered set.

I disagree. The above interval is a perfectly well defined interval.
Its end points (in general) represent two non-equatorial points of
a parallelized 3-sphere. The even sub-algebra of the algebra Cl(3,0)
is a well ordered set, homeomorphic to a parallelized 3-sphere,
which, in turn, is a well defined topological space, endowed with
a well defined order topology. You are not going to catch me out
on some technicality.

> >                   sigma(A)  =  [ d f / d w ] x sigma(w)
>
> >                                  =  v x sigma(w).
>
> Since your interval is not well defined, this is not a valid
> inference.

Since my interval is well defined, my inference is perfectly valid.

Joy Christian

[Heine Rasmussen]

On 16 Feb, 22:49, Joy Christian <hojoin...@gmail.com> wrote:
> On Feb 16, 4:59 pm, Heine Rasmussen <hein...@gmail.com> wrote:
>

[...]

>
> > Intervals have no meaning in GA exept for the scalar subspace. A
> > Clifford algebra is not a well ordered set.
>
> I disagree. The above interval is a perfectly well defined interval.
> Its end points (in general) represent two non-equatorial points of
> a parallelized 3-sphere. The even sub-algebra of the algebra Cl(3,0)
> is a well ordered set, homeomorphic to a parallelized 3-sphere,
> which, in turn, is a well defined topological space, endowed with
> a well defined order topology. You are not going to catch me out
> on some technicality.

So now you are claiming that S^3 has an order topology? That's
impossible, since it is a well known theorem that any separable,
compact, connected space with an order topology is homeomorphic to the
unit interval [0,1] of reals.

Nor is it possible to give any sensible total order relation on the
even sub-algebra Cl(3,0).

Heine Rasmussen

[Joy Christian]

Please read carefully what I have written. I have written
"...parallelized 3-sphere." You are not going to catch me
out on some technicality. You are not the first one to try.

Joy Christian

[Heine Rasmussen]

And how does that matter? A "parallelized 3-sphere" is still not
homeomorphic to the unit interval, is it?

Basing an argument om an interval devoid of any meaning, whis leads to
a conclusion that trivially violates a topological theorem is hardly
"some technicality".

Heine Rasmussen

[Joy Christian]

On Feb 18, 6:22 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> And how does that matter?  A "parallelized 3-sphere" is still not
> homeomorphic to the unit interval, is it?

That is a homework for you to find out.

The interval I considered earlier, namely

[ m(w) - sigma(w), m(w) + sigma(w) ],

is a perfectly well-defined interval on a
parallelized 3-spehre. Your arguments
therefore have no mathematical basis.

Joy Christian

[Heine Rasmussen]

On 18 Feb, 20:03, Joy Christian <hojoin...@gmail.com> wrote:
> On Feb 18, 6:22 pm, Heine Rasmussen <hein...@gmail.com> wrote:
>
> > And how does that matter?  A "parallelized 3-sphere" is still not
> > homeomorphic to the unit interval, is it?
>
> That is a homework for you to find out.

The question was rethorical; we all know it's not homeomorphic to the
unit interval.

> The interval I considered earlier, namely
>
>   [ m(w) - sigma(w), m(w) + sigma(w) ],
>
> is a perfectly well-defined interval on a
> parallelized 3-spehre. Your arguments
> therefore have no mathematical basis.

There are no references to multivector intervals in the literature.
It is easy to see that your derivation only goes through for v and w
both grade-0, but then the standard derivation is a scalar, anyway.
So the concept of a bivector valued standard deviation is still as
unsupported as ever.

Heine Rasmussen

[Joy Christian]

On Feb 19, 7:07 pm, Heine Rasmussen <hein...@gmail.com> wrote:

> The question was rethorical; we all know it's not homeomorphic to the
> unit interval.

The question was also irrelevant for the present discussion.

> There are no references to multivector intervals in the literature.
> It is easy to see that your derivation only goes through for v and w
> both grade-0, but then the standard derivation is a scalar, anyway.
> So the concept of a bivector valued standard deviation is still as
> unsupported as ever.

I beg to differ. In addition to the extensive discussion in my paper
(which is self-contained), I have given you several clear-cut (albeit
simplified) arguments to show how bivector valued standard
deviations can arise naturally within a parallelized 3-sphere (which
is the basis for my local-realistic model). To this, you objected by
claiming that the interval I considered in one of my arguments

earlier, namely

[ m(w) - sigma(w), m(w) + sigma(w) ],

is ill-defined. I disagree with your claim. Therefore it is now your
turn to demonstrate (in comparable details to what I have provided)
how and why the above interval is ill-defined, and moreover tell us
why this ill-definedness is relevant -- if at all -- for my model. So
far you have only made unsubstantiated claims (if not false).

Joy Christian