According to the following paper published yesterday the much hyped "loophole free" experiment done by the Delft group and discussed in this blog is deeply flawed because it violates the no-signaling condition respected by quantum mechanics:
I would advise to think twice before trying to "spook."
John R. Cox replied on Nov. 15, 2015 @ 16:44 GMT
Tom,
I'm out if my depth here, but can recognize that what is at the bottom of the issue is the difference inherent in the time parameter between relativistic time and quantum time. Quantum time normalizes to 1 because 1sec/1sec = 1sec. But that is the same as Newtonian time and goes to instantaneous measure because the only 1sec it can refer to is an ad hoc arbitrary earth second. And that omits that time is in motion also. The problem mathematically that seems to confront the logician, is the incompleteness of arithmetic; we can multiply by zero but not divide by zero. So a dynamic (relativistic) time parameter encounters the problem in expression of 1sec/0sec = 'error' even though physically the result should be 1sec that is not divided is still 1 second, and 1sec that is wholly divided is zero.
Physically, at the beginning of a measure of motion the time parameter is zero but at the end of a measure of motion the time parameter is at 1. Time has been in motion as well as the object moving in that time, and Quantum time does not recognize that. Normalization is stagnant time because of the lack of completeness in arithmatic that prompted George Boole's attempt to rationalize in algebraic logic.
"We find once again, the failure to account for the behavior of time is..." THR
The behavior of time part, I get. :-) jrc
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Thomas Howard Ray replied on Nov. 15, 2015 @ 21:52 GMT
I get that you don't get it, Rick.
Alice and Bob don't have free choice of a and b simultaneously.. The function is continuous -- when the limit on one of the variables is reached, the other is compelled to reverse sign, and the function continues.
In fact, one need only understand two things to understand Joy's framework -- initial condition, and continuous function on a simply connected space. You and your compatriots are not even playing ball in the same park.
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Thomas Howard Ray replied on Nov. 16, 2015 @ 14:27 GMT
John,
As interesting as your argument is, I don't want to get the issue of time reversibility -- key to the Hess-Phillip and Bednorz programs -- confused with Joy Christian's program.
In the former, time is a
physical parameter. In the latter, it is an
extra-physical parameter. In other words, Joy makes the time dimension equal to the space by using a topological framework of added dimension.
The result is the same -- time reversibility is the same as conservation of spin angular momentum " ... EPR-Bohm correlations are correlations among the scalar points of a quaternionic 3-sphere."
It's frustrating that the critics have not been able to see this -- covering their mistakes with irrelevant technical arguments that do not apply to the measure space.
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Rick Lockyer replied on Nov. 16, 2015 @ 15:12 GMT
Tom, it is not a lack of understanding on my part, it is completely on you.
You wrote: "Alice and Bob don't have free choice of a and b simultaneously.. The function is continuous -- when the limit on one of the variables is reached, the other is compelled to reverse sign, and the function continues."
Besides the fact that this is complete nonsensical gibberish, it is not Joy's idea of a and b. Quote, page 6 of his book second edition: "In what follows we shall view the fixed bivectors (-I.a) and (+I.b) as representing the measurement instruments....". Initial conditions you seem so keen on, wouldn't you think??
"You and your compatriots are not even playing ball in the same park. "
Damn straight, one of a small list of things we agree on. You see, I demand mathematical correctness within any physical model that I will deem credible. Not sure who my "compatriots" are, but I do not pretend to speak for anyone but myself.
"It's frustrating that the critics have not been able to see this -- covering their mistakes with irrelevant technical arguments that do not apply to the measure space. "
No, what is frustrating is that the proponents are so enamored by the story line they can't accept the very relevant technical argument that mathematics does not support the story. The proponents are the ones covering their mistakes in a rather juvenile fashion. Not sure you have your audience characterized here. I think it is clear to most that the critics are not the ones making the mistakes. Your obfuscations are completely transparent, and the math errors pointed out by the critics are quite trivial when one cuts through the "you just do not understand" BS.
Your pet phrases: continuous functions, simply connected space are descriptive, not causal. You put mystical properties on them that only exist between your ears.
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Thomas Howard Ray replied on Nov. 16, 2015 @ 19:33 GMT
Rick,
(quoting Joy) "In what follows we shall view the fixed bivectors (-I.a) and (+I.b) as representing the measurement instruments....". Initial conditions you seem so keen on, wouldn't you think??"
No. The text follows, " ... for detecting the random bivectors (+ mu.a) and (+ mu.b) which represent the spins." The paragraph following:
"It is crucial to note that the variables A(a, lambda) and B(b, lambda) are generated with
different bivectorial scales of dispersion (or different standard deviations) for each direction a and b."
It should be quite obvious to the informed reader that you take the initial condition to be a detector setting while Joy takes it as a state of spacetime.
And this is a pattern with you and other critics -- build a straw man, knock it down, and then preen over your vast mathematical knowledge, while denigrating attempts to set you straight as nonsense. One must be willing to learn.
"Your pet phrases: continuous functions, simply connected space are descriptive, not causal. You put mystical properties on them that only exist between your ears."
Entanglement is not causal, either. In the case of a continuous measurement function over a simply connected manifold, however, the assumption that measure space precedes measure
is causal.
If you reply with "nonsensical gibberish" I won't have anything more to say to you.
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Rick Lockyer replied on Nov. 16, 2015 @ 22:35 GMT
Tom, straw man defense for an indefensible position. Nothing new here eh? You have yourself in a hole, stop digging.
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John R. Cox replied on Nov. 17, 2015 @ 16:05 GMT
In a topological space, as distinct from an infinite 2D analytical plane, the change of sign is incorporated in the quaternion algebra which makes possible the determination of points as position within a 3D octant of a sphere but without the inversion of 'up or down' orientation of the point object when projecting into another octant, as can occur on a complex plane in projection from one quadrant into another.
Interpreting results as if S1 is a complex plane interior of a circle seems to be how arguments for spookiness picture the plot map of correlated pairs. On a complex plane the time parameter has to flip-flop across an axis into negative territory but in topology the direction of time is the same in all octants. "time reversibility is the same as conversion of spin angular momentum"
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Fred Diether replied on Dec. 1, 2015 @ 07:20 GMT
@Rick Lockyer replied on Nov. 16, 2015 @ 22:35 GMT
Lockyer got himself into a big hole over not being able to tell the mathematical difference between a left handed system and a right handed system from a right handed perspective. But I see he still keeps on digging. It is so freakin' simple...
For a pair of particles that are in a singlet configuration, Nature has a 50-50 chance that they will be created as either a left handed system or right handed system. And that is just plain common sense. For a left handed system the expression is;
(I.a)(I.b) = -a.b - (a^b) = -a.b - I.(a x b) (1) LH System
For the right handed system we have;
(I.a)(I.b) = -a.b - (a^b) = -a.b - I.(a x b) (2) RH System
Note that both equations are the same. However, eq. (1) is in a left handed basis and eq. (2) is in a right hand basis so we can't add the left hand expression to the right hand expression properly. Now, translate eq. (1) to the right hand basis so that we can do that.
But first I want to stress, "However, eq. (1) is in a left handed basis and eq. (2) is in a right hand basis so we can't add the left hand expression to the right hand expression properly." The translation of eq. (1) to the right hand basis is simply;
(I.b)(I.a) = (-I.a)(-I.b) = -a.b - (-I).(a x b) = -a.b + a^b
GAViewer cofirms that this is correct.
-a.b + a^b
ans = 0.77 + 0.39*e2^e3 + -0.44*e3^e1 + -0.25*e1^e2
(I.b)*(I.a)
ans = 0.77 + 0.39*e2^e3 + -0.44*e3^e1 + -0.25*e1^e2
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Rick Lockyer replied on Dec. 1, 2015 @ 19:37 GMT
Fred, apparently you do not believe (+1)(+1) = (-1)(-1). This is elementary school mathematics. I told you before your error is as simple as this.
You wrote: "(I.b)(I.a) = (-I.a)(-I.b) = -a.b - (-I).(a x b) = -a.b + a^b"
The minus signs on the I terms are identical to multiplication by the real number -1, and since it is a real number, it commutes with any other product independent of the algebra that product represents. Thus the two minus signs cancel and you are left with (I.b)(I.a) = (I.a)(I.b). But (I.a) and (I.b) are bivectors in a non-commutative algebra, so clearly you have made a mistake.
You, Joy Christian and Tom Ray need to apologize to a long list of people that have tried too hard in my opinion to teach you what you should already know.
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Fred Diether replied on Dec. 4, 2015 @ 23:07 GMT
Fine. Just remove (-I.a)(-I.b) if it is bothering you. The fact remains that,
(I.b)(I.a) = -a.b - (-I).(a x b) = -a.b + a^b,
is true. And it is the proper translation of eq. (1) above for the LH (left hand) system to the right hand perspective so one can add them together. So now there should be no problem with the math of Joy's model. It works as originally advertised to produce the result of -a.b for the EPR-Bohm scenario. No apology necessary. Perhaps you would like to apologize.
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Rick Lockyer replied on Dec. 5, 2015 @ 03:54 GMT
For the last time Fred:
(I.a) and (I.b) are bivectors represented in one handedness based on the definition of I and the arbitrary right vs. left chiral designation. Then (-I.a) and (-I.b) would be the same bivectors represented in the opposite handedness. By what you have just agreed on then we have
(I.a)(I.b) = (-I.a)(-I.b)
This is precisely what I have been saying. Chiral choice for the algebra is arbitrary, it is a symmetry that when properly represented yields a totally equivalent result. The non scalar portion of the GA product in Joy's flawed math does not sum to zero since the left and right handed forms are equivalent. Stop being stuck on st...d.
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Fred Diether replied on Dec. 5, 2015 @ 07:27 GMT
Now you are in a hole so deep you might as well keep digging because you will never get out. I told you to forget about (-I.a)(-I.b) since it bothers you so much. It is not the important part. The important part is and always will be,
-a.b - (-I).(a x b) = -a.b + a^b
Now, if you think that you can add terms from a left hand basis to terms in a right hand basis without translating one of them to the other basis, then keep on digging. It is really pathetic that you are not able to see that you lost this debate and you want to concentrate on something that doesn't matter at all. There is no flaw in Joy's math and there never has been.
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Rick Lockyer replied on Dec. 5, 2015 @ 16:31 GMT
Fred wrote "Now, if you think that you can add terms from a left hand basis to terms in a right hand basis without translating one of them to the other basis, then keep on digging."
Having memory issues or just a disconnection from reality? This was my statement to you when you first tried to justify JC's position on your own, It was precisely what you were doing and the only way to come to the incorrect conclusions you have. I DID the derivation for you explicitly using the left and right handed bivector basis elements and you said there were no GA errors but refused to admit it proved there are math errors. That position itself demonstrated your lack of mathematical sophistication. You did it again here not being able to see the explicit basis element derivation is identical to the (+1)(+1)=(-1)(-1) argument.
This is not a debate, it is very simple mathematics you lack the ability to understand and apply. What is your next move Fred, your "difference between math and a physics postulate" nonsense?
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John R. Cox replied on Dec. 5, 2015 @ 16:59 GMT
"GA product" refers specifically to a topological app developed as 'Graphic Arts Viewer'. It operates on only one of the two possible 'handedness' settings, to have both right and left handed differentiated locations plotted against an ideally spherical surface requires incorporating a sidewise sign change.
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Rick Lockyer replied on Dec. 5, 2015 @ 17:24 GMT
GA here is Geometric Algebra
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Thomas Howard Ray replied on Dec. 5, 2015 @ 18:26 GMT
Rick, does your mathematical sophistication permit you to understand what a quaternionic 3-sphere is?
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Joy Christian replied on Dec. 5, 2015 @ 23:14 GMT
What "sophistication", Tom? Lockyer's gobbledygook is not worth paying any attention to, as
Michel Fodje pointed out long time ago.
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Thomas Howard Ray replied on Dec. 6, 2015 @ 03:41 GMT
Ah, yes, Joy. I remember that post! Powerful words, and solid advice. Thanks to you and Michel.
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Fred Diether replied on Dec. 6, 2015 @ 16:42 GMT
Yeah, it looks like Michel got the incompetent part right. I found Lockyer's sophomoric math error in his
presentation here. A clue is that it stands to common sense logic that if the formula for left and right handed systems are the same at the start, then when one is translated to the other that they will not be equal to each other. For lurkers that might want to discover Lockyer's mistake for themselves, another clue can be found in the
multiplication tables here.. In his presentation, only the C and F calculations matter. F is wrong for another clue.
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Rick Lockyer replied on Dec. 7, 2015 @ 04:50 GMT
Sorry guys but there are no math errors in any of my criticisms. It is one thing to make a non-specific claim of an error and quite another to explicitly state what is wrong and what correct is. Step up, do it.
Fred, you are so ignorant about algebra it is a waste of effort to reply. There is no basis for comparison until you cast the two handedness representations in the same basis, your last post was quite telling. Tom, you live in an alternate universe where mysticism trumps logic. Joy, you really ought to know better.
Any of you fools. Fodje included, explicitly tell us all how the following is not correct:
Define "I" such that for the GA vectors a and b the bivectors (I.a) and (I.b) are "right handed". Then bivectors (-I.a) and (-I.b) would be "left handed". This is straight out of Joy's declarations, as anyone that takes the Fodje link Joy put up in this thread can see, so hard to imagine any of you will dispute this without some laughable argument.
But...
(-I.a)(-I.b) = {(-1)(I.a)}{(-1)(I.b)} = (-1)(-1)(I.a)(I.b) = (I.a)(I.b)
right handed == left handed
No BS now, no fake straw-man crap. Where is the error?? This is as easy as it gets. Anyone that follows the Fodje link Joy put up can see he does not think there is an equality.
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Joy Christian replied on Dec. 7, 2015 @ 06:07 GMT
Tom, Fred, please ignore the noise generated by Lockyer by respecting the first rule of Internet. Let the non-fool believe whatever he wishes to believe.
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Fred Diether replied on Dec. 7, 2015 @ 06:20 GMT
Well, I thought I would give Lockyer some time to find his math error in the presentation linked above and even gave some clues to finding it. I will post the solution tomorrow. It is pretty simple. Not sure why he hasn't discovered it after all these months. Or I guess he is so arrogant that he probably didn't even look for his error. In the meantime, perhaps he can answer the question about how if the formula for left and right handed systems start out the same (equal), how could they possibly be equal after one of them is translated to the other perspective. It is just not possible. It defies common sense logic. That little fact by itself should motivate him or anyone to look for his error since he had them equal before and after translation.
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Anonymous replied on Dec. 7, 2015 @ 14:17 GMT
Rick, though I should follow Joy's sage advice, maybe there's a learning opportunity here.
Fred is right. Translation extends domain.
Special relativity, for example, depends on Lorentz transformation to keep its relation between points constant. Otherwise, one lives with the illusion of having objects appear in two places, or multiple places, at the same time (Einstein lensing).
Once one understands the nature of spacetime, one understands the illusion.
It's the same with left hand and right hand algebraic bases. A simple translation from one to the other changes the rules -- from a duality of bases to a single base of common spacetime. This is possible, because spacetime was never truly divided. When you say -- "There is no need for the limits s->a and s->b. The claim A and B are = +/-1 and they are the product of two bivectors imply these pairs are parallel or anti-parallel, not in the limit. The dummy variable s in s->a is not the same dummy variable s in s->b, since Alice and Bob have free choice on a and b. It is blatantly obvious then that one cannot assume the two L's in s can be combined." -- you arbitrarily divide the choice function into a duality of Alice and Bob.
This arbitrary division keeps the RH and LH algebras independent, but it yields no insight except as you say, the "blatantly obvious." When one realizes that Alice and Bob cannot choose their a's and b's at the same time, the analogy with special relativity should be clear. One does need the limits, and the calculus subsumes the algebra.
That is why I say that neither Bell nor Gill eliminated time from physics. The algebra did.
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Thomas Howard Ray replied on Dec. 7, 2015 @ 14:19 GMT
Rick Lockyer replied on Dec. 7, 2015 @ 18:02 GMT
Tom, look at the math instead of your own ethereal thoughts on the subject. The two L's in question can ONLY be combined if Alice's s is ALWAYS the same as Bob's s. Clear math error. Get your head out of the clouds.
Fred, perhaps you are missing the obvious fact that we are dealing with the product of two bivectors each of which is represented in the same right or left handed system. Think about it.
Joy, the best you can do?
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Fred Diether replied on Dec. 7, 2015 @ 18:33 GMT
The post exposing Lockyer's math error is
here. From his comment above to Tom, it seems he still does not "get it" even after giving him a ton of clues. He made the simple sophomoric error of double mapping F thus there was no translation done at all in his calculation for F. As can be seen at the link above, it proves that Joy's model is correct a little bit different way.
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Thomas Howard Ray replied on Dec. 7, 2015 @ 20:39 GMT
Rick, my head's more comfortable in the clouds, than in the sand.
Fred, excellent.
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Fred Diether replied on Dec. 7, 2015 @ 21:09 GMT
Tom, LOL! and thanks. It might be interesting to see if Lockyer tries to wiggle his way out of this since there is no wiggle room left for him.
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Thomas Howard Ray replied on Dec. 7, 2015 @ 21:40 GMT
Fred, the problem is flatland thinking.
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Fred Diether replied on Dec. 7, 2015 @ 21:59 GMT
Well... to be fair to Lockyer, for him it was a case of self-brainwashing. He was so convinced that there was no error in the presentation he did it resulted in his self-brainwashing conviction that Joy had to be wrong no matter what. It will also be interesting if any apologies are forthcoming from him. But I am not going to hold my breath waiting for such.
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Rick Lockyer replied on Dec. 7, 2015 @ 22:23 GMT
There is a block copy editing error only on the last sum line for C,D,E and F, did not change some 3's to 2's. The simplifications that follow immediately have no errors, so sorry but there is no math error.
C =
+ (a_0 b_0) e_0*e_0 + (a_1 b_1) e_1*e_1 + (a_2 b_2) e_2*e_2 + (a_3 b_3) e_3*e_3
+ (a_0 b_1) e_0*e_1 + (a_1 b_0) e_1*e_0 + (a_2 b_3) e_2*e_3 + (a_3 b_2)...
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There is a block copy editing error only on the last sum line for C,D,E and F, did not change some 3's to 2's. The simplifications that follow immediately have no errors, so sorry but there is no math error.
C =
+ (a_0 b_0) e_0*e_0 + (a_1 b_1) e_1*e_1 + (a_2 b_2) e_2*e_2 + (a_3 b_3) e_3*e_3
+ (a_0 b_1) e_0*e_1 + (a_1 b_0) e_1*e_0 + (a_2 b_3) e_2*e_3 + (a_3 b_2) e_3*e_2
+ (a_0 b_2) e_0*e_2 + (a_2 b_0) e_2*e_0 + (a_3 b_1) e_3*e_1 + (a_1 b_3) e_1*e_3
+ (a_0 b_3) e_0*e_3 + (a_3 b_0) e_3*e_0 + (a_1 b_2) e_1*e_2 + (a_2 b_1) e_2*e_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (a_0 b_1 + a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (a_0 b_2 + a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (a_0 b_3 + a_3 b_0 + a_1 b_2 – a_2 b_1) e_3
D =
+ (a_0 b_0) e_0*f_0 + (a_1 b_1) e_1*f_1 + (a_2 b_2) e_2*f_2 + (a_3 b_3) e_3*f_3
+ (a_0 b_1) e_0*f_1 + (a_1 b_0) e_1*f_0 + (a_2 b_3) e_2*f_3 + (a_3 b_2) e_3*f_2
+ (a_0 b_2) e_0*f_2 + (a_2 b_0) e_2*f_0 + (a_3 b_1) e_3*f_1 + (a_1 b_3) e_1*f_3
+ (a_0 b_3) e_0*f_3 + (a_3 b_0) e_3*f_0 + (a_1 b_2) e_1*f_2 + (a_2 b_1) e_2*f_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (– a_0 b_1 + a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (– a_0 b_2 + a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (– a_0 b_3 + a_3 b_0 – a_1 b_2 + a_2 b_1) e_3
E =
+ (a_0 b_0) f_0*e_0 + (a_1 b_1) f_1*e_1 + (a_2 b_2) f_2*e_2 + (a_3 b_3) f_3*e_3
+ (a_0 b_1) f_0*e_1 + (a_1 b_0) f_1*e_0 + (a_2 b_3) f_2*e_3 + (a_3 b_2) f_3*e_2
+ (a_0 b_2) f_0*e_2 + (a_2 b_0) f_2*e_0 + (a_3 b_1) f_3*e_1 + (a_1 b_3) f_1*e_3
+ (a_0 b_3) f_0*e_3 + (a_3 b_0) f_3*e_0 + (a_1 b_2) f_1*e_2 + (a_2 b_1) f_2*e_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_0 b_1 – a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (a_0 b_2 – a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (a_0 b_3 – a_3 b_0 – a_1 b_2 + a_2 b_1) e_3
F =
+ (a_0 b_0) f_0*f_0 + (a_1 b_1) f_1*f_1 + (a_2 b_2) f_2*f_2 + (a_3 b_3) f_3*f_3
+ (a_0 b_1) f_0*f_1 + (a_1 b_0) f_1*f_0 + (a_2 b_3) f_2*f_3 + (a_3 b_2) f_3*f_2
+ (a_0 b_2) f_0*f_2 + (a_2 b_0) f_2*f_0 + (a_3 b_1) f_3*f_1 + (a_1 b_3) f_1*f_3
+ (a_0 b_3) f_0*f_3 + (a_3 b_0) f_3*f_0 + (a_1 b_2) f_1*f_2 + (a_2 b_1) f_2*f_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (– a_0 b_1 – a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (– a_0 b_2 – a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (– a_0 b_3 – a_3 b_0 + a_1 b_2 – a_2 b_1) e_3
No change: C=F, and when added, the non-scalar term does not go to zero. This typo was pretty damn obvious, don't you think? Let me spell out the simplifications for you:
f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 = e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3
You can see from the e and f indexes how the signs on the bivector results will be the same for right and left handed representations when cast in the e basis. But then this is more simply represented by
(I.a)(I.b) = (-I.a)(-I.b)
Sorry for the typo, but it was not a supportive argument for you anyway. The first 3 sums resulting in e_0, e_1 and e_2 were correct as they stood. This is extremely easy and straight forward, not my problem the three of you can't do simple algebra.
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Fred Diether replied on Dec. 7, 2015 @ 22:37 GMT
Amazing! After his double mapping error is explicitly pointed out to him, Lockyer is still in denial. And he did the double mapping again above. Well I tried to help him out as anyone reading here can witness for themselves. I guess we are done here since there is no hope for him to be able to do simple algebra correctly.
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Rick Lockyer replied on Dec. 7, 2015 @ 23:26 GMT
Fred, you really need to take an elementary algebra class, get a book or something. The non-scalar results for basis element products in the left handed (f) and right handed (e) multiplication tables are accurately indicated in
f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 = e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3
You can't add an e system coefficient to an f system coefficient, both values added must first be cast in the same basis, then and only then can they be added. I am guessing this is your obtuse and totally bonehead reference to double translation. This is algebra 101. Like I said before not my problem you lack the ability to see it.
Seriously, think about what the equality (I.a)(I.b)=(-I.a)(-I.b) means. The issue at hand REALLY is this simple. This equation would seem to be more your speed. Why don't you try to refute it.
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Fred Diether replied on Dec. 8, 2015 @ 01:10 GMT
Quite amazing that you can't see that the mapping is already in the tables. So go ahead and hit the calculation with the mapping again. It is pretty plain to see that the mapping is already incorporated in the tables. Are you just plain dense? Double up on it. We really don't care what you do anymore. You are hopeless.
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Rick Lockyer replied on Dec. 8, 2015 @ 03:22 GMT
Fred, baby steps just for you with a_0 = b_0 = 0.
First lets agree e and f as defined correctly represent right and left handed systems respectively, and the map between is negation of the non-scalar basis elements. If you disagree then you must tell us which of the following is not correct:
f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 = e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3
Work through it Fred, do the e_x -> -f_x and f_x -> -e_x substitutions in the above to verify the map works. Compare the basis products to some multiplication table source you trust. If there is a problem, point it out, else man up and say it is correct.
Taking the above product definitions, we have for a_0 = b_0 = 0
F =
+(a_1 b_1) f_1*f_1 + (a_2 b_2) f_2*f_2 + (a_3 b_3) f_3*f_3
+(a_2 b_3) f_2*f_3 + (a_3 b_2) f_3*f_2
+(a_3 b_1) f_3*f_1 + (a_1 b_3) f_1*f_3
+(a_1 b_2) f_1*f_2 + (a_2 b_1) f_2*f_1
=
-(a_1 b_1 + a_2 b_2 + a_3 b_3) f_0
+ (–a_2 b_3 + a_3 b_2) f_1
+ (–a_3 b_1 + a_1 b_3) f_2
+ (–a_1 b_2 + a_2 b_1) f_3
=
-(a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_2 b_3 - a_3 b_2) e_1
+ (a_3 b_1 - a_1 b_3) e_2
+ (a_1 b_2 - a_2 b_1) e_3
Now compare this to C with a_0 = b_0 = 0 already provided. Same-oh Jack. So if I am wrong, the error is somewhere just above, and you should be able to tell us all exactly what is wrong and what correct is. Put up, or shut up and stop spamming the internet with ignorant BS.
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Fred Diether replied on Dec. 8, 2015 @ 04:09 GMT
If you can't see that the mapping is already in the tables, there is not much I can do to help you. Go ahead and just keep double mapping the calculation. The following is wrong,
-(a_1 b_1 + a_2 b_2 + a_3 b_3) f_0
+ (–a_2 b_3 + a_3 b_2) f_1
+ (–a_3 b_1 + a_1 b_3) f_2
+ (–a_1 b_2 + a_2 b_1) f_3
The mapping was already done by the multiplication in the left hand table. It should read,
-(a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (–a_2 b_3 + a_3 b_2) e_1
+ (–a_3 b_1 + a_1 b_3) e_2
+ (–a_1 b_2 + a_2 b_1) e_3
Get yourself some help for this if you don't know how to do the tables correctly. Stop with your double mapping already.
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Rick Lockyer replied on Dec. 8, 2015 @ 04:28 GMT
Sorry Fred, but your post was non-responsive. I showed you each step and ask you to identify the error, you have not done this. You made a blanket statement which certainly is what you wish it was since you have gone way over the line of acceptable discourse with many critics with it when you are clearly wrong.
Show your work, disclose the full f multiplication table (f only chump) and the full e multiplication table (e only) and your map between. Apply them and show how you get to the conclusion you did. Otherwise explicitly tell me which step(s) I did were wrong and justify your position. Ask Joy and Tom for help if you must, but show the work as I have if you can't explain which steps of mine are incorrect.
The nonsense coming out of the three of you has to stop.
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Fred Diether replied on Dec. 8, 2015 @ 04:50 GMT
You
are dense. Sorry, can't help you out of your quagmire. Michel was certainly right. Bye.
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Joy Christian replied on Dec. 8, 2015 @ 05:48 GMT
Fred, in addition to what you say, Rick Lockyer is also dishonest. His math errors have been pointed out to him literally hundreds of times, here at FQXi and elsewhere, but he dishonestly disregards every opportunity to recognize them. It is also worth noting that he has no qualifications in physics to understand something as basic as what is meant by a "hidden variable", or what is the actual physics behind the EPR-Bohm type experiments. All of these things contribute to his ignorance and inability to recognize his math errors.
By contrast, my 3-sphere calculations have been explicitly verified by Lucien Hardy, a FQXi member and one of the foremost experts in the foundations of quantum mechanics. Moreover, one of
my papers on the arXiv where I show this calculations has been
published in the International Journal of Theoretical Physics, whose founding editor as well as several of its distinguished editorial board members are also members of FQXi.
To be sure, one does not have to depend on any such "argument from authority." All one has to do is to look at just a few lines of my elementary calculations in
my latest paper (for example) to verify for themselves that my mathematical and physical arguments are trivially correct.
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Luca Valeri replied on Dec. 8, 2015 @ 09:05 GMT
Hope I can ask a question without getting insulted. But please do not stop here, as I followed the discussion for a while and you all really seem to have boiled it down to simple arithmetics.
If a basis change from left to right handed is done by f_1 -> - e_1 for the first component, then
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
looks true to me. But then the first component from C and F are the same. Or did I miss something?
Best regards
Luca
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Thomas Howard Ray replied on Dec. 8, 2015 @ 14:12 GMT
That's the point, Luca. Simple arithmetic can't capture the effect of the hidden variable. Substitution is accomplished "by hand," as if the elements were independent.
Such introduces a duality, where none exists. The algebra is correct, following the rules of algebra. However, a simple translation from one base to the other changes the rules -- from a duality of bases to a single base of common spacetime. This is possible, because spacetime was never truly divided.
To take no account of limits, is to assume that the domain is flat Euclidean. I wish that I could answer your question more directly -- however, I can't until you specify domain and limit.
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Rick Lockyer replied on Dec. 8, 2015 @ 16:58 GMT
Luca, thanks for stepping up. You are missing nothing, the left and right handed system multiplications and mapping between have been accurately presented by me and DO show ALL components of C and F are the same when either represented in the right OR the left handed bases. This demonstrates to a mathematical certainty only one of a plethora of math errors Joy Christian has in his work
You also were spot on when you said it is simple math, and I do hope you don't get personally attacked by Joy Christian, Fred Diether, or Tom Ray. Seems to be where they go when called on inaccurate positions by someone who wants the truth to prevail. Calling it juvenile is too kind. It should ONLY be about the math.
Joy, you had a good start and novel idea in the beginning. Too bad you jacked it all up with bad math. Stay with the S^n program but abandon the current tack you are on, it is clearly a non-starter. Look closer at the physics of the particle and measurement. Only things have meaningful handedness differences, for spaces it is a free choice for representation, a symmetry when anything measurable is accurately represented within it.
You keep bringing up my "lack of qualifications in physics". Normally this is when your back is up against a wall on the math, a position you find yourself in all too frequently. For me math is totally unfulfilling without physics, and I am not about to turn in my PHYSICS degree from Stanford any time soon.
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Thomas Howard Ray replied on Dec. 8, 2015 @ 17:57 GMT
"Only things have meaningful handedness differences, for spaces it is a free choice for representation ..."
And that is where you err.
Since my own explanations of failure to include spacetime are "mystical" to you ... perhaps you will pay due attention to
Andrei Khrennikov:"Bell’s theorem [3] states that there are quantum spin correlation functions that can not be represented as classical correlation functions of separated random variables. It has been interpreted as incompatibility of the requirement of locality with the statistical predictions of quantum mechanics [3]. For a recent discussion of Bell’s theorem see, for example [4] - [18] and references therein. It is now widely accepted, as a result of Bell’s theorem and related experiments, that ” local realism” must be rejected and there exists the so called quantum non-locality.
However it was shown in [17, 18, 20] that in the derivation of such a conclusion the fundamental fact that space-time exists was neglected. Moreover, if we take into account the spatial dependence of the wave function then the standard formalism of quantum mechanics might be consistent with local realism."
One
does not have free choice of representation for spacetime.
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Fred Diether replied on Dec. 8, 2015 @ 18:08 GMT
Hi Luca,
Thanks for your question. The following expression is not correct.
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
It should be,
f_2 * f_3 = -e_1 (left hand)
and e_2*e_3 = +e_1 (right hand)
So it is easy to see that the translation mapping is already done. What Lockyer did is to hit -e_1 with the mapping again to give +e_1 in his calculation. IOW, he double mapped (converted) the calculation. For sure C will equal F when you do that since no translation was ever done.
Another note is that f_1 -> - e_1 does not mean exactly that f_1 is equal to -e_1. It depends on what perspective you are "viewing" from.
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Rick Lockyer replied on Dec. 8, 2015 @ 23:35 GMT
Tom, we are talking about orientation choice here. Admittedly I skimmed the article you referenced, but I saw nothing about orientation.
Fred, you continue to demonstrate your ignorance of algebra. The two algebras e and f are self contained, meaning the product of two f basis elements is another f, not an e. Now you can always define a map between any two (quaternion == bivector) systems, which is precisely what I have done. The maps are not unique, but all are isomorphic. What this means is if you have some other, the end result will be the same mod basis choice and its attendant mapping. Put another way, if you have shown one, you have demonstrated all.
I have given you the e and f system basis element products relevant to orientation, and the map between which completes the picture of their orientation difference. I invited you to look closely at the two systems to verify they indeed have opposite orientation and the mapping between is consistent, and when the algebras are correctly applied, you only get opposite sign coefficients when one is in the e basis and the other is in the f basis. The isomorphism seals the deal on this always being the case, and like I have said since day one, you can't add coefficients unless the basis elements attached are identical.
You will be stuck on stupid until you sit down and look at what I have provided with the intent to understand it. I think you can understand it, but won't try because you will not like the steady diet of crow that you will realize you have brought on yourself.
Stop spamming FQXi, you are behaving like a troll. You have already damaged your reputation enough.
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Joy Christian replied on Dec. 9, 2015 @ 00:04 GMT
Thanks again, Fred. Your reply to Luca makes Lockyer's sophomoric mistake quite transparent for everyone to see. Since he has been making his silly mistake for so many years now, it is unlikely that he will ever admit to having made it. It is truly mindboggling where his breathtaking arrogance stems from, considering the fact that he can't even do simple algebra correctly.
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Fred Diether replied on Dec. 9, 2015 @ 00:51 GMT
You're welcome, Joy. Another mystery is why Lockyer is so confused about what the multiplication tables are doing.
Luca, if the tables are confusing you also, just ask and I will do a longer more detailed explanation. Use
this web page as a reference for any questions.
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Rick Lockyer replied on Dec. 9, 2015 @ 02:14 GMT
Nothing new in both of your last responses. It is really too bad you can't just do the mathematical analysis to justify your claims that your detractors have ACTUALLY made a mistake. I defined the e and f algebras short of the products that never change like scalar * not scalar and like not scalar all with well known results. In other words I have fully defined each. I provided the map between e and f short of the obvious e_0 = f_0 = 1. I applied in baby steps with nothing omitted what amounts to the product of two like coefficient bivectors using both e and f algebras, and then applied the map between algebras to the results. These results justify the claims by Christian's detractors that he made a sign error.
Clearly you do not like the results. But all you have done is make claims that I like all the detractors have made a mistake, we do not know what we are talking about, sophomoric errors, etc. etc. Words only, no math. All that you have needed to do to support your claim of accuracy is back up from my results you do not like to the step(s) where I have made a mistake and point it out. Every step is there, nothing is left to imagination or interpretation.
Why have you not done this? Seems like a better approach than empty personal attacks. My opinion: Joy already has and knows there are no errors, but has gone too far down the road he is on to admit it. My opinion on Fred: apparently clueless.
Do the math, show your work as I have so we can all put this to bed one way or the other.
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Fred Diether replied on Dec. 9, 2015 @ 02:27 GMT
Luca Valeri replied on Dec. 9, 2015 @ 08:43 GMT
Thanks Fred for the reply and for the useful link to the multiplication tables.
As I read them, this means that for any right handed basis you have
r_2 * r_3 = r_1
For any left handed basis you have:
l_2 * l_3 = - l_1
So the e basis with e_2 * e_3 = e_1 is right handed, and the left handed basis f must have
f_2 * f_3 = - f_1
Right? So what is the step to get:
f_2 * f_3 = - e1 ?
Thanks
Luca
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Joy Christian replied on Dec. 9, 2015 @ 10:14 GMT
I am not amused by Lockyer's empty rhetoric and relentless threats. I have endured much online bullying and harassment from the likes of him over the past several years.
Lockyer's sophomoric math errors have been exposed, literally hundreds of times, in many different ways, with explicit mathematical steps, and in published papers. See for example the link just posted by Fred. See also my replies in these papers:
(1)(2)(3)These are just a sample of
explicit mathematical refutations I have provided.
Luca,
please read in full the link Fred has just provided, including my re-post of a post there.
As Fred has explained, Lockyer's mistake is quite easy to see. He has double translated in going from left to right to get C = F, thus making a sign error. It is as simple as that. Quite easy to see in the link posted by Fred.
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Luca Valeri replied on Dec. 9, 2015 @ 12:14 GMT
Hi Joy, Fred
I tried to follow Fred's link. But I'm not interested to follow Lockyer's mistake. I'd like just to do baby steps, so I can follow your and/or Fred's arguments. So what is wrong with f_2*f_3=-f_1 for a left handed basis? And how do to derive f_2*f_3=-e_1, where e_1 is a basis vector of the right handed basis?
Luca
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Thomas Howard Ray replied on Dec. 9, 2015 @ 13:18 GMT
Rick, it should be obvious that insofar as spacetime is a "thing" with handedness properties, the article is all about orientation.
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Joy Christian replied on Dec. 9, 2015 @ 14:26 GMT
Luca,
It is quite simple.
As explained
here by me (in GA notation), and as noted also by Fred above, both left and right handed basis satisfy EXACTLY THE SAME EQUATION. Therefore we have
f_2 * f_3 = + f_1 (left hand)
and
e_2 * e_3 = + e_1 (right hand).
Now do your mapping,
f_1 -> - e_1 ,
to get
f_2 * f_3 = - e_1 ,
where e_1 is a basis vector of the right handed basis (note that this looks awkward only because of the non-GA notation used here -- it is much more clear in the above
link).
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Rick Lockyer replied on Dec. 9, 2015 @ 17:03 GMT
Luca wrote:
"As I read them, this means that for any right handed basis you have
r_2 * r_3 = r_1
For any left handed basis you have:
l_2 * l_3 = - l_1
So the e basis with e_2 * e_3 = e_1 is right handed, and the left handed basis f must have
f_2 * f_3 = - f_1
Right? So what is the step to get:
f_2 * f_3 = - e1 ?
Thanks
Luca...
view entire post
Luca wrote:
"As I read them, this means that for any right handed basis you have
r_2 * r_3 = r_1
For any left handed basis you have:
l_2 * l_3 = - l_1
So the e basis with e_2 * e_3 = e_1 is right handed, and the left handed basis f must have
f_2 * f_3 = - f_1
Right? So what is the step to get:
f_2 * f_3 = - e1 ?
Thanks
Luca "
Yes Luca you have this correct and what Christian just put up is nonsense. Handedness is defined by the signs of the basis element products, not something you anoint as he just did to two identically signed expressions in the same indexes but different alphabetic labels.
The left and right algebras stand on their own as algebras, meaning all basis element products must produce another in the set. To compare the two algebras e and f here there must be some commonality between them. What is employed here is the indexes. So here we have the three coefficients for a and b making the common connection between the e basis and f basis. So with commonality through the index assignments if we have e_2*e_3=e_1 we must have f_2*f_3=-f_1 if e and f have opposite handedness. Christian is backing into the results he wants by bluffing, something he unfortunately does a lot.
One cannot add Ae_1 to Bf_1 even with the common connection provided by the same index. It is only possible to add AFTER translation of e_1 to f_1 or f_1 to e_1. So there must be some mapping provided. What is required from this map is to be able to substitute the e basis equivalents into the f multiplication table and reproduce the e multiplication table, AND substitute the f basis equivalents into the e multiplication table and reproduce the f multiplication table. Keeping index commonality, the only possible mapping is
e_0 < - > f_0
e_1 < - > -f_1
e_2 < - > -f_2
e_3 < - > -f_3
If you make these substitutions either way in the following multiplications
f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 = e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3
you end up with A=A, B=B..... So what is the conclusion so far? The answer is the e and f algebras are correctly defined from above with the additional always the case products e_0*e_j = e_j = e_j*e_0 and e_j*e_j = -e_0 for j 1 to 3, and with f exactly the same. The mapping e < - > f works both ways.
So next we do the math as I did in both e and f algebras, then map the f result to the e basis and compare it to the e basis result. No surprise to anyone understanding what an automorphism is, the result is the same.
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Fred Diether replied on Dec. 9, 2015 @ 17:13 GMT
Hi Luca,
Ok, let me explain in more detail. Lockyer submitted the following table,
f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 = e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3
Which reduces to;
f_1 * f_2 = e_1 * e_2
f_2 * f_3 = e_2 * e_3
f_3 * f_1 = e_3 * e_1
f_2 * f_1 = e_2 * e_1
f_3 * f_2 = e_3 * e_2
f_1 * f_3 = e_1 * e_3
IOW, the left hand system is equal to the right hand system after translation of the left to right. It is
absurd!. If that is true, why even bother with the multiplication tables? It is easy to see that the left hand table is already converting to the right hand basis. Thus Lockyer's mistake in not recognizing that.
It is easier to visualize what is going on by simply using the cross product in vector analysis than by doing what Lockyer is doing. Imagine that we have two vectors a and b at right angle to each other. Their cross product results in a third vector orthoganal to the plane that a and b form, a x b = c. Now if you look up the left and right hand rules you will find that if a is pointing up and b is pointing out of the page then for a right hand system, c will point to the right and for a left handed system, c will point to the left. So what does the left handed system look like from the right hand perspective? c is pointing the other way from the right hand c so it must be -c in the right hand perspective. Thus the left hand system translated to the right perspective is b x a = -c. The order of multiplication is reversed. This proves that a left handed system will not be equal to a right handed system after translation. Thus Lockyer made a mistake.
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Thomas Howard Ray replied on Dec. 9, 2015 @ 17:17 GMT
"The left and right algebras stand on their own as algebras ..."
To the limit of the domain.
One of the conclusions in the Khrennikov/Volovich paper linked earlier allows, "One can get quantum nonlocality in the EPR situation only if we rather artificially restrict ourself in the measurements with a two dimensional subspace of the infinite dimensional Hilbert space corresponding to the position or momentum observables."
Still want to continue down this road?
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Joy Christian replied on Dec. 9, 2015 @ 17:26 GMT
What Lockyer just put up is pure obfuscation -- a lame attempt to hide his sophomoric sign mistake. As I have already noted, his mistake is quite easy to see. He is cheating by doing double translation.
As explained
here by me (in GA notation), and as noted also by Fred above, both left and right handed basis satisfy EXACTLY THE SAME EQUATION. Therefore we have
f_2 * f_3 = + f_1 (left hand)
and
e_2 * e_3 = + e_1 (right hand).
Now simply do the mapping,
f_1 -> - e_1 ,
f_2 -> - e_2 ,
f_3 -> - e_3 ,
to get
e_2 * e_3 = - e_1 (left hand, from the perspective of right hand),
where e_i are basis vectors of the right handed basis. This is clearly explained also at the above
link, as well as in many of
my papers. Lockyer is simply cheating.
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John R. Cox replied on Dec. 9, 2015 @ 17:43 GMT
Fred,
Thank-you for the clarity of presentation in your simple paragraph giving visualization to the geometry of the algebra, I found that generously easy to follow. In the same simply stated way, might you elaborate on *why* the reversal of multiplication (in non-commutative norm) "proves" chiral inequality in translation across the minus divide? Thanks, again, jrc
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Fred Diether replied on Dec. 9, 2015 @ 18:10 GMT
Hi John,
Thanks for your question. b x a = -c is just a standard identity in vector analysis. You should be able to find it doing a search. But think about what the left handed system looks like from the right hand perspective. If you look hard enough, you will realize that the multiplication is indeed reversed. It is a pure physical thing.
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Eckard Blumschein replied on Dec. 9, 2015 @ 19:09 GMT
John C,
I hope you didn't need Fred's hint to "a standard identity in vector analysis". Otherwise I would like asking you to imagine three orthogonal defined as positive directions in space: For instance: velocity of a moving conductor (a) causes by mediation of a magnetic field (b) an induced voltage (c). This is the so called right-hand rule with your thumb (a), index finger (b), and middle finger (c).
Imagine (a) rotated within the area ab to (b). This gives a positive sign to the area ab with (c) a vector directed orthogonal to it. Rotation from (b) to (a) corresponds to a view on the area ba which points in the opposite (negative) direction. Screws are usually right-wound.
++++
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Rick Lockyer replied on Dec. 9, 2015 @ 20:46 GMT
Tom, not my road to continue down. The correctness of Bell's theorem is neither here nor there as far as my points here go.
On your comment "Rick, it should be obvious that insofar as spacetime is a "thing" with handedness properties, the article is all about orientation." I would think within an article "all about orientation" one could search the document for the word "orientation" and get at least one hit. More connections that only exist in the world of "Tom".
This article does nothing to support the claim by Christian there are no math errors in his work, which after all IS what this thread has digressed to.
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Rick Lockyer replied on Dec. 9, 2015 @ 21:01 GMT
Fred, the chiral change map is an automorphism. Please go away and learn what that means.
Joy, normally you do a better job of hiding your nonsense, but
"
f_2 * f_3 = + f_1 (left hand)
and
e_2 * e_3 = + e_1 (right hand).
"
is so obviously wrong I can't imagine how you think you could possibly pass this off.
Wow!
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Fred Diether replied on Dec. 9, 2015 @ 21:19 GMT
Lockyer's math error exposed.
I doubt very much that Lockyer could provide an online example of this so-called automorphism between left handed and right handed physical systems. It doesn't exist.
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Joy Christian replied on Dec. 9, 2015 @ 21:26 GMT
Let me note one more time to bring the point home. Lockyer is either cheating, or is genuinely clueless about what is being discussed in my papers. Anyone serious enough about physics is welcome to check out my argument presented at
this link, or in some of my papers:
(1)(2)(3)As I noted above, these are just a handful of my papers where I have
explicitly refuted misguided arguments like that of Lockyer.
Let me also quote Michel Fodje's response to Lockyer's bogus claims:
"There is no sign error. If you do not see what has been explained to you yet then it is you who is incompetent. If you do see it but can't admit your error, then you are being disingenuous. Your arguments on this thread are not about Joy's model, but rather a figment of your imagination."
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Rick Lockyer replied on Dec. 9, 2015 @ 23:03 GMT
Joy wrote:
"Now simply do the mapping,
f_1 -> - e_1 ,
f_2 -> - e_2 ,
f_3 -> - e_3 , "
Fred wrote:
"Which reduces to;
f_1 * f_2 = e_1 * e_2
f_2 * f_3 = e_2 * e_3
f_3 * f_1 = e_3 * e_1
f_2 * f_1 = e_2 * e_1
f_3 * f_2 = e_3 * e_2
f_1 * f_3 = e_1 * e_3
IOW, the left hand system is equal to the right hand system after translation of the left to right. It is absurd!. If that is true, why even bother with the multiplication tables? It is easy to see that the left hand table is already converting to the right hand basis. Thus Lockyer's mistake in not recognizing that."
So Fred, do Joy's substitutions. Like I told you earlier, you must not believe that (+1)(+1)=(-1)(-1). Want to revise your position here??
So Joy, use your own substitutions on your correct and undisputed by me claim
"e_2 * e_3 = + e_1 (right hand)."
=> (-1)f_2 * (-1)f_3 = (-1)f_1
=> f_2 * f_3 = -f_1
How do you square this with your follow on
"f_2 * f_3 = + f_1 (left hand)
and
e_2 * e_3 = + e_1 (right hand)."
Care to revise your position here Joy??
None of this came from me fools. It all came from you. This does not even rise to "school boy howler" status. Breathtakingly stupid.
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Joy Christian replied on Dec. 9, 2015 @ 23:17 GMT
Lockyer is cheating once again, or is genuinely clueless about what is presented in my papers.
He is equating heads with tails of a fair coin,
heads = tails ,
and then crying wolf.
His purpose clearly is to hoodwink the scientific community by this deception.
I, for one, am not fooled by his paper wolf.
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Fred Diether replied on Dec. 9, 2015 @ 23:21 GMT
Yeah, and I am still waiting for Lockyer to produce this so-called automorphism example. He can't do it because it doesn't exist.
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John R. Cox replied on Dec. 10, 2015 @ 04:06 GMT
Thank-you Fred and Eckard, visualizing handedness similar to the taper of a common wood screw as alagorical to the leftward ascending values on the initial condition 'minus' side, and physically reversing the direction of torsion on that side to mimic an opposing torsion of the initial minus axis, is (I think) an essential pedagogical 'wall chart' necessary to instruct readers as to the prescribed protocols that come under discussion.
Whether called 'the leap-frog effect' of electromagnetic induction, 'the right hand rule', and the arbitrarily assigned 'positive' : 'negative' polarity and hence 'north' : 'south'; it is always so easily explosive into that nagging doubt, "is it flopped now here? or have I flipped out?" Without continual grounding in the physical realm topologically, any one oversight in one parameter while progressing through a protocol can easily be obfuscated in the otherwise raw of math. jrc
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Eckard Blumschein replied on Dec. 10, 2015 @ 05:52 GMT
Rick,
You wrote: "the chiral change map is an automorphism." May I naively translate automorphism in this case as perfect symmetry between past and future? This could be water on my mills.
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Joy Christian replied on Dec. 10, 2015 @ 07:08 GMT
Here is a
counterexample to Lockyer's uninformed claim that "the chiral change map is an automorphism." This counterexample is well known to any physicist since 1956. He has a habit of insulting Fred and Tom on regular basis, with breathtaking delusions about his own knowledge and competence.
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John R. Cox replied on Dec. 10, 2015 @ 15:08 GMT
P conservation would be illustrated by a standard barrel nut, threaded continuo